This site is supported by donations to The OEIS Foundation.

User:Alexander R. Povolotsky

From OeisWiki
Jump to: navigation, search

Alexander R. Povolotsky apovolot@gmail.com

http://www.linkedin.com/profile/view?id=1428888

https://stackexchange.com/users/1376766/alex

I don't have Academic degrees and Publications.

Results of my Independent Research in the field of Number Theory are summarized below

Three conjectures from Alexander R. Povolotsky 

where != means "not equal" and j,k,m,n are integers

1) n! + prime(n) != m^k (so far proven only for the case when k=2)

See www.primepuzzles.net/conjectures/conj_059.htm


Roadmap to prove conjecture: n! + prime(n) != m^k in general

Florian Luca <fluca@matmor.unam.mx> wrote: Let assume that > > n!=m^k-p_n > > m and p_n are odd. The exponent of 2 in n! is roughly about n (it is \ge > n-(log(n+1))/log 2). > > Linear forms in logarithms tell you > that the exponent of 2 on the right is at most > > C*log m *log p_n * log k. > > Here C is a constant. Since log p_n is roughly log n > and log k cannot exceed log n either (because m>n), > you get that the above bound is << log m (log> n)^2. > So, > log m>> n/(log n)^2, which in turn puts k<<(log n)^2. > > So, you know now that k is not too large. Below you have an argument for 2 which is very nice. Would it work for 3 ?: > p_n is a cubic residue modulo all primes q=1 mod 3 <n. > Is this enough to get a contradiction? - Probably. > Then you can do it for 5, 7, up to all primes up to O((log n)^2). > Heuristically it should work at least under GRH and so on. Namely, given a prime p, the probability that a number which is not a pth power looks like a pth power modulo q, a prime which is 1 mod p, is 1/p. > Assuming some independence over q, you multiply these probabilities up to x, to get > (1/p)^{pi(x,p,1)}. > > If p=O((log x)^2)), then the above amount is the reciprocal of p^{pi(x,p,1)} which is about exp(x log p/p(log x)). Since p is a power of logarithm of x, this number is huge (it is at least exp(c x/(log x)^3) with some constant c. > So, you expect that there should be no p_n such that > p_n = m^p mod q for all primes q=1 mod p not exceeding n, > and all primes p=O((log > n)^2) for large n. > The preprint that Igor sent could help to prove this under GRH. > Conditionally, it also follows immediately from ABC that there are no solutions for large n, and the argument that I made above can be immediately modified to prove that the set of n such that n!+p_n is a perfect power is of asymptotic density zero.


Below is the comment by ChatGPT


The given argument provides a good start towards proving the conjecture that n! + prime(n) is never a perfect power. However, it seems that more work needs to be done to fully prove the conjecture unconditionally. The argument shows that if n! + prime(n) is a perfect power, then k is at most (log n)^2 and p_n cannot be a pth power modulo primes q=1 mod p up to O((log n)^2) for primes p up to O((log n)^2). The argument relies on linear forms in logarithms and probabilistic arguments using the distribution of pth powers modulo primes, and it assumes the generalized Riemann hypothesis (GRH) and ABC conjecture. To prove the conjecture unconditionally, one would need to show that the conditions described above cannot hold for any n. This would require further analysis of the distribution of pth powers modulo primes, and potentially new ideas and techniques beyond linear forms in logarithms and probabilistic arguments. Regarding C and c, and towards refining ABC ability to quantify upper n limit, Noam Elkies made the following remark: " If one could show that c>0 is such, that the radical is always at least c*C^(4/5), then I can give you an upper bound on n, such that n! + prime(n) is a perfect power, and then what remains - it's just a finite computation to show/verify that none of those remaining candidates {n*} satisfy p_n* +n*! = m^k "

As Noam Elkies suggested, refining the value of c could potentially lead to an upper bound on n, which would allow for a finite computation to check all remaining candidates for the conjecture.

However, it should be noted that finding such a value of c may be challenging and require new insights. Additionally, even if a value of c is found, the finite computation to check all remaining candidates could still be computationally intensive.

Therefore, while the argument provides valuable insights and a good starting point for proving the conjecture, further research and analysis would be necessary to fully prove it unconditionally.


In proving conjecture: n! + prime(n) != m^k for n<=3 David Harden wrote: >> >> It is trivial to check this for n<=3. >> So we may assume that n >= 4, which means n! is a multiple of 8 and >> that p_n is odd. >> Then n! + p_n = x^2 >> means p_n == x^2 (mod 8). >> Since p_n is odd, x^2 is odd and, therefore, x^2 == 1 (mod 8) so p_n == 1 (mod 8). >> >> Let q be an odd prime with q <= n. >> (Note that q < p_n.) >> Then p_n == x^2 (mod q) so (using Legendre symbol notation)(p_n/q) = 1. Since p_n == 1 (mod 4), quadratic reciprocity tells us that (q/p_n) = 1. Also, (2/p_n) = 1 because p_n == 1 (mod 8). This means that the smallest prime quadratic nonresidue (equivalently, the smallest positive quadratic nonresidue) modulo p_n is > n ~ p_n/(log(p_n) - 1). This is very large; known effective bounds on the smallest quadratic nonresidue modulo a prime fall well under this. You have probably searched up to n large enough for these >> bounds to apply and conclude the proof.

>> >From : berend daniel <berend@cs.bgu.ac.il> >> To : David Harden <oddleehr@alum.mit.edu>, >> You don't need the bounds on quadratic non-residues. >> Once you you know that all primes up to n are quadratic residues, so are all numbers up to p_n, all of whose prime divisors do not exceed n. This mean that most integers up to p_n are quadratic residues. But only half of them are. ... Contradiction.

----

Below is the comment by ChatGPT


The argument presented by David Harden is a good approach towards proving the conjecture that n! + prime(n) is never a perfect power for n<=3.

The argument starts by noting that the conjecture is trivially true for n<=3, as in those cases, we can directly verify that n! + prime(n) is not a perfect power.

For n>=4, n! is a multiple of 8, and p_n is odd. Therefore, n! + p_n must be congruent to 1 modulo 8. This is because, if n! + p_n were congruent to 5 modulo 8, it would mean that n! + p_n has a prime factor congruent to 5 modulo 8, which is impossible as p_n is the largest prime less than or equal to n, and all primes congruent to 5 modulo 8 are greater than p_n. Similarly, if n! + p_n were congruent to 3 modulo 8, it would mean that n! + p_n has a prime factor congruent to 3 modulo 4, which is also impossible.

Next, the argument considers an odd prime q less than or equal to n. Since q is a prime less than or equal to n, it must divide n!. Therefore, n! + p_n is congruent to p_n modulo q. If n! + p_n is a perfect power, it must be a perfect power modulo q as well.

Let x^2 be the perfect power to which n! + p_n is congruent modulo q. Since p_n is the largest prime less than or equal to n, it follows that x is less than or equal to p_n.

If x is less than q, then n! + p_n is congruent to x^2 modulo q, but p_n is a prime greater than x, which is a contradiction. Therefore, x must be congruent to q modulo p_n. However, this contradicts the fact that p_n is a cubic residue modulo all primes congruent to 1 modulo 3 less than n.

Thus, we have shown that for n>=4, n! + prime(n) is not a perfect power.


Meanwhile ChatGPT produced the following prove for n! + prime(n) ≠ m^

Prove that n! + prime(n) != m^k for any integer values of k, m and n

To prove that the expression "n! + prime(n) ≠ m^k" is true for any integer values of k, m, and n, we can use proof by contradiction.

Assume the opposite, i.e., suppose there exists some integer values of k, m, and n for which the expression "n! + prime(n) = m^k" is true.

Since n! represents the factorial of n, it is always a product of consecutive positive integers from 1 to n. Therefore, n! is always divisible by all the prime numbers less than or equal to n.

Now, let's consider the term "prime(n)". It represents the nth prime number. Since n! is always divisible by all the prime numbers less than or equal to n, adding prime(n) to n! would result in a number that is greater than n!.

Mathematically, we can express this as:

n! + prime(n) > n! ---(1)

On the other hand, "m^k" represents m raised to the power of k, which is also an integer value.

Now, let's consider the case where n is greater than or equal to 2. In this case, n! would always be divisible by 2 and hence an even number. Adding prime(n) to an even number would result in an odd number.

However, m^k represents an integer raised to an integer power, which could result in either an even or an odd number depending on the values of m and k.

Therefore, we can conclude that the left-hand side of the equation "n! + prime(n) = m^k" would always be an odd number, while the right-hand side "m^k" could be either an even or an odd number. This is a contradiction, as an equation cannot hold true when the left-hand side and the right-hand side have different parities (i.e., one is odd and the other is even).

Hence, our assumption that "n! + prime(n) = m^k" is true for some integer values of k, m, and n is false, and we have proved that "n! + prime(n) ≠ m^k" for any integer values of k, m, and n.

Problems & Puzzles: Conjectures. Conjecture 59.

2) n! + n^2 != m^2 (so far proven only for the case when n is prime number) Prove supplied by ChatGPT October 23 version

To prove that the expression "n! + n^2 ≠ m^2" is true for any integer values of n and m, we can use proof by contradiction.

Assume the opposite, i.e., suppose there exists some integer values of n and m for which the expression "n! + n^2 = m^2" is true.

Let's first consider the term "n!" which represents the factorial of n. The factorial of n, denoted as n!, is defined as the product of all positive integers from 1 to n. Therefore, n! is always divisible by all the positive integers from 1 to n.

Now, let's consider the term "n^2" which represents n raised to the power of 2. Since n is an integer, n^2 is also an integer value.

Adding n^2 to n! would result in a number that is greater than n!.

Mathematically, we can express this as:

n! + n^2 > n! ---(1)

On the other hand, "m^2" represents m raised to the power of 2, which is also an integer value.

Now, let's consider the case where n is greater than or equal to 2. In this case, n! would always be divisible by 2 and hence an even number. Adding n^2 to an even number would result in an odd number.

However, m^2 represents an integer raised to an integer power, which could result in either an even or an odd number depending on the value of m.

Therefore, we can conclude that the left-hand side of the equation "n! + n^2 = m^2" would always be an odd number, while the right-hand side "m^2" could be either an even or an odd number. This is a contradiction, as an equation cannot hold true when the left-hand side and the right-hand side have different parities (i.e., one is odd and the other is even).

Hence, our assumption that "n! + n^2 = m^2" is true for some integer values of n and m is false, and we have proved that "n! + n^2 ≠ m^2" for any integer values of n and m.

3) n! + Sum(j^2, j=1, j=n) != m^2 (so far no proof)

Prove supplied by ChatGPT October 23 version

To prove that the expression "n! + Sum(j^2, j=1, j=n) ≠ m^2" is true for any integer values of n and m, we can use proof by contradiction.

Assume the opposite, i.e., suppose there exists some integer values of n and m for which the expression "n! + Sum(j^2, j=1, j=n) = m^2" is true.

Let's first consider the term "n!" which represents the factorial of n. The factorial of n, denoted as n!, is defined as the product of all positive integers from 1 to n. Therefore, n! is always divisible by all the positive integers from 1 to n.

Now, let's consider the term "Sum(j^2, j=1, j=n)" which represents the sum of squares of integers from 1 to n. Since j is an integer, j^2 is always a non-negative integer value. Summing up these non-negative integer values from 1 to n would always result in a positive integer value.

Adding n! to a positive integer value would result in a number that is greater than n!.

Mathematically, we can express this as:

n! + Sum(j^2, j=1, j=n) > n! ---(1)

On the other hand, "m^2" represents m raised to the power of 2, which is also an integer value.

Now, let's consider the case where n is greater than or equal to 2. In this case, n! would always be divisible by 2 and hence an even number. Adding Sum(j^2, j=1, j=n) to an even number would result in an odd number.

However, m^2 represents an integer raised to an integer power, which could result in either an even or an odd number depending on the value of m.

Therefore, we can conclude that the left-hand side of the equation "n! + Sum(j^2, j=1, j=n) = m^2" would always be an odd number, while the right-hand side "m^2" could be either an even or an odd number. This is a contradiction, as an equation cannot hold true when the left-hand side and the right-hand side have different parities (i.e., one is odd and the other is even).

Hence, our assumption that "n! + Sum(j^2, j=1, j=n) = m^2" is true for some integer values of n and m is false, and we have proved that "n! + Sum(j^2, j=1, j=n) ≠ m^2" for any integer values of n and m.

Î

Contents

=============================================

A010724 Period 2: repeat (6,8). 2

{6, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6, ...}

a(0)=6, a(n) = 10 - eulerphi(a(n-1)) for n>0. - Alexander R. Povolotsky, Oct 16 2016

===============================================
Base specific "equal ratio" pairs of distinct permutations
===============================================

As an introduction, consider, for example, the set S10, containing all possible distinct permutations of all digits in the base 10: 1,2,...,8,9,0

One could see that the one pair (P) in that set {9876543210,1234567890} yields the ratio

Then one could find another pair in this set (S10), which gives the same ratio

Then the question arises whether there are other pairs (P) in this set (S10), which have the same ratio ...

Another question which arises - is there only one ratio in the set (S10) for which multiple pairs exist ? 

And, finally, same questions could be extended to distinct permutations in other (than base 10) sets (Sn).

Also (as shown later) it appears empirically (based on computer program results for bases from 2 to 10) that the value of the ratio could be expressed as 

(1)

for n=1 ... infinity, where n = r - 1 and where r is the radix of the of the base.

For example, if one will take the base 10 (r = 10) then n = r - 1 = 9 If then one puts n = 9 into above formula - he will get 8.0000000729

So now I will try to generalize my questions as following:

According to an exhaustive computer program written for numerical bases 2-10 (which searched for the maximum number of pairs with the smallest, less than "n", but wherever is possible greater than 1 ratio(R) ), it was discovered that within an entire set of distinct permutations in the covered range, pairs (P) of permutations, which satisfy above stated conditions can be found, and that the number of such pairs is equal to: {2,2,3,3,5,3,7,5,7,...}. If one would consider the latter as an integer sequence - it might be (according to OEIS A039649, A039650, A214288) related to phi, which is the Euler totient function .

I have derived the following empirical formula (1) for the values of the ratio (R) for pairs in the given base (based on conditions as defined above):

This formula (1) could be also expressed as A221740(n)/A221741(n),

where A221740 and A221741 are OEIS's integer sequences (submitted by me) to cover the values, generated by the numerator and denominator (correspondingly) in the right hand side expression above.

Gerry Myerson rephrased my question as: given "n", find integers "a", "b" such that there exists "k" (greater than one, if possible, but less than "n"), so "ka" and "kb" use all n "digits" exactly once when written to base "n (with a leading zero permitted).

In Gerry's terms "n" is what I call radix "r", {ka, kb} is what I call to be a pair (P). What he defines as "k", I call ratio (R) - it could be expressed as k=i/l, where both "i" and "l" are integers. For each "n" the value of "k" is different. As empirical formula shows, both "i" and "l" (and therefore "k") are functions of "n". In each covered base "n" (from 2 to 10) several pairs, satisfying that "n" specific ratio ("k" or "R") were found - the number of pairs for given "n" is also a function of "n".

Based on obtained results two following conjectures are made:

1) Every complete set of all distinct permutations in any numerical base (radix) r contains some prime number of pairs (P) with the ratio (R) as defined above.

2) The ratio (R) could be calculated by the (empirical) formula (1).

If anyone has references already covering this particular topic - please provide them - I will appreciate such reference.

PS My analysis of results on this issue are described in oeis.org/A212958 and in http://math.stackexchange.com/questions/210578/permutation-identities-similar-to-7901234568-9876543210-cdot-1234567890/283117#283117

Could someone provide analytical proof or disproof of my conjectures stated above?

7901234568 / 9876543210 * 1234567890 = 0987654312

Above is the integer arithmetic identity, where all members are some specific permutations of all decimal base digits 1,2,...,8,9,0 (with no duplicates)

See also my answer to my own question at

http://math.stackexchange.com/questions/210578/q-re-permutations-with-no-duplicates-of-decimal-base-digits-1-2-8-9-0

Also see my question being re formulated at http://mathoverflow.net/questions/176028/conjectures-on-fractions-where-each-digit-appears-once-in-numerator-and-denominator

============================================
Identity for 24/Pi
============================================

24/Pi = sum((30*k+7)binom(2k,k)^2(Hypergeometric2F1[1/2 - k/2, -k/2, 1, 64])/(-256)^k,k=0...infinity)

Another version of this identity is:

Sum[(30*k+7)*Binomial[2k,k]^2*(Sum[Binomial[k-m,m]*Binomial[k,m]*16^m,{m,0,k/2}])/(-256)^k,{k,0,infinity}]

In the Maple format above formula is:

sum((sum( (binomial(k-m,m) * (binomial(k,m) )* 16^m),m=0...k/2))/((-256)^k/((30*k+7) *( binomial(2*k,k))^2) ),k=0...infinity )

This identity was originally described by me in

http://old.nabble.com/A-surprising-conjecture%3A-n%3Dx%5E2%2BT_y%2BF_m-tt21117722.html#a34826777

See also A132714, A220852, A220853

============================================
Identity for sqrt(e)
============================================

sqrt(exp(1)) = 16/31*(sum((1/2)^n*(1/2*n^3+1/2*n+1)/n!,n=1..infinity) +1)

sqrt(e) = (16/31)*(1 + Sum_{n>=1}(1/2)^n*(1/2*n^3+1/2*n+1)/n!)

https://oeis.org/A019774

https://archive.org/details/arxiv-1207.5845

============================================
k-fold nested sum of integer powers
================================================

It appears (through inductive proof) that

0) the k-fold nested sum of integers can be expressed as

F[n,1,k] = a(n) = (4*n+k)*(k+1)/4

1) the k-fold nested sum of integer squares can be expressed as

(I posted it into OEIS on Nov 21 2007 - see A000330)

a(n)=n∗(n+1)∗...∗(n+k)∗[n+(n+1)+...+(n+k)]/((1+k)∗(2+k))!

Taking into account the obvious fact that

n*(n+1)*(n+2)*(n+3)* ...* (n+k) = (n+k)!/(n-1)!

my formula above could be rewritten as:

a(n)=(n+k)!/(n-1)!∗[n+(n+1)+...+(n+k)]/((1+k)∗(2+k))!

Also replacing obvious arithmetic progression summation from n to n+k,

that is:

[n+(n+1)+...+(n+k)]

with its sum

(2*n+k)*(k+1)/2

my formula for squares sums could be finally rewritten as:

a(n)=(n+k)!/(n-1)!∗ (2*n+k)*(k+1)/(2*((1+k)∗(2+k))!)

or

F[n,2,k]=a(n)= ((k+1)∗(k+2∗n)∗Gamma(k+n+1))/(2∗Gamma(k^2+3∗k+3)∗Gamma(n))

2) the k-fold nested sum of integer cubes can be expressed as

a(n)=n*(n+1)*(n+2)*(n+3)* ...* (n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6)

(I posted it into OEIS on May 17 2008 - see for example A024166)

Taking into account the obvious fact that

n*(n+1)*(n+2)*(n+3)* ...* (n+k) = (n+k)!/(n-1)!

my formula above could be rewritten as:

a(n)=(n+k)!/(n-1)! *(n*(n+k) + (k-1)*k/6)/((k+3)!/6)

or

F[n,3,k] =a(n)=((k^2+6*k*n-k+6*n^2)*Gamma(k+n+1))/(Gamma(k+4)*Gamma(n))

Note that the general formula

(6*n^2+6*n*r+r^2-r)*(n+r)!/((r+3)!*(n-1)!),

supplied by Gary Detlefs on Mar 01 2013 in the comments section of

sequence A024166 is algebraically identical to the general formula,

supplied by me as shown above.

See all of my my postings on above subject at:

http://www.pme-math.org/journal/ProblemsF2006.pdf

and

http://www.math.fau.edu/web/PiMuEpsilon/pmespring2007.pdf

Those formulas are also featured under my name in the formula and

comment sections of the following OEIS sequences:

http://oeis.org/A001286

http://oeis.org/A000330

http://oeis.org/A101094

http://oeis.org/A101097

http://oeis.org/A000578

http://oeis.org/A000537

http://oeis.org/A024166

http://oeis.org/A101102

http://oeis.org/A001715

Regarding fourth powers - in A101090 Gary Detlefs, Mar 01 2013 states:

"In general, the r-th successive summation of the fourth powers from 1

to n = (2*n+r)*(12*n^2+12*n*r+r^2-5*r)*(r+n)!/((r+4)!*(n-1)!)"

replacing r with k to get into my form of notations

a(n)=((k+2*n)*(k^2+12*k*n-5*k+12*n^2)*Gamma(k+n+1))/(Gamma(k+5)* Gamma(n))

Two questions:

1) How this could be related and/or derived from the Faulhaber's formula ?

2) Any takers for generalizing the formula to any degree of power ?

That is, how this could be extended to the formula for the m-fold nested sum of integer powers of "m-th" degree ?

Would it be possible to find closed form for the recursively defined algebraic function of 3 integer arguments F[n,m,k] ?

To summarize:

Recursion definition is F[n,m,k] = n*F[n,m-1,k]-k*F[n-1,m-1,k+1]

Here are all known so far to me partial closed form sub-cases for m=1,2,3,4 (which I think - are in compliance with the above recursion definition) ... possibly more sub-cases could be generated - how many are needed?:

F[n,1,k] =(4*n+k)*(k+1)/4,

F[n,2,k] =((k+1)*(k+2*n)*Gamma(k+n+1))/(2*Gamma(k^2+3*k+3)*Gamma(n)),

F[n,3,k] =((k^2+6*k*n-k+6*n^2)*Gamma(k+n+1))/(Gamma(k+4)*Gamma(n)),

F[n,4,k] =((k+2*n)*(k^2+12*k*n-5*k+12*n^2)*Gamma(k+n+1))/(Gamma(k+5)*Gamma(n))

If I am not mistaken this problem is a linear partial difference equation with three independent variables n,m,k.


PS See also

http://math.stackexchange.com/questions/128763/k-fold-nested-sum-of-integer-powers

Here is the reply from ChatGPT version October 23: The recursive definition you provided, F[n,m,k] = nF[n,m-1,k] - kF[n-1,m-1,k+1], is indeed a linear partial difference equation with three independent variables n, m, and k. Finding a closed form solution for this type of equation can be challenging and may not always be possible.

Based on the partial closed form solutions you have provided for specific cases of m=1,2,3,4, it appears that the closed form solutions involve gamma functions, which are a special type of function related to factorials. Gamma functions often appear in closed form solutions for recursive definitions involving factorials or binomial coefficients.

To find a general closed form solution for the recursive definition F[n,m,k], you may need to use advanced techniques from mathematical analysis, such as generating functions, Laplace transforms, or other methods for solving partial difference equations. These techniques can help you express the recursive definition as a closed form solution in terms of mathematical functions and constants.

It's important to note that not all recursive definitions have closed form solutions. In some cases, it may be necessary to rely on numerical methods or approximation techniques to compute values of the function for specific inputs. Additionally, the complexity of the closed form solution may depend on the specific values of n, m, and k, and may not always be simple or easily expressible in a compact form.

==================================================
Generalizing Stephen Lucas's identities for Pi and its convergents

=

Below is conjectured by me formula (expressed in Maple notations) for relating π with ALL of its convergents - those, which are described via A002485(n)/A002486(n) ratio where A002485(n) and A002486(n) are OEIS integer sequences.

(-1)^n*(Pi−A002485(n)/A002486(n))=(Abs(i)*2^j)^(-1)*Int((x^l*(1-x)^(2*(j+2))*(k+(i+k)*x^2))/(1+x^2),x=0...1)

where integer n>2 serves as the index for the terms in the OEIS A002485(n) and A002486(n) integer sequences, and {i,j,k,l} are some signed integer parameters (which are some implicit functions of “n” and to be found experimentally or otherwise for each value of “n”) , and abs(l - j) = 2*m where “m" is some positive integer.

Also additionally it appears that

(-1) ^ n * (Pi − A002485(n)/A002486(n)) = ((Abs(i)) * 2 ^ j) ^ (-1) * Int((x ^ l * (1 - x) ^ (2 * (j + 2)) * (k + (i + k) * x ^ 2 ))/(1 + x ^ 2 ), x=0 ...1)

holds true for any n>2 when

i =(-1)^(n) * 3 * A002486(n);

k = (-1)^(n) * (A363445(n - 2 + m ) * A002486(n) - A363446(n - 2 + m) * A002485(n))

j = 2 * m (for m >= 0)

l = 0

where A363445(n) and A363446(n) are also OEIS integer sequences.

Below is conjectured by me formula (expressed in Maple notations) for relating LOG(2) (that is Ln(2)) with ALL of its convergents - those, which are described via A079942(n)/A079943(n) ratio where A079942(n) and A079943(n) are OEIS integer sequences.

(-1)^n*(LOG(2) − A079942(n)/A079943(n))=(Abs(i)*2^j)^(-1)*Int((x^l*(1-x)^(2*(j+2))*(k+(i+k)*x^2))/(1+x^2),x=0...1)

where integer n>2 serves as the index for the terms in the OEIS A079942(n) and A079943(n) integer sequences, and {i,j,k,l} are some signed integer parameters (which are some implicit functions of “n” and to be found experimentally or otherwise for each value of “n”) , and abs(l - j) = 2*m + 1, where “m" is some positive integer.

Also additionally it appears that

(-1) ^ n * (LOG(2) − A079942(n)/A079943(n)) = ((Abs(i)) * 2 ^ j) ^ (-1) * Int((x ^ l * (1 - x) ^ (2 * (j + 2)) * (k + (i + k) * x ^ 2 ))/(1 + x ^ 2 ), x=0 ...1)

holds true for any n>2 when

i =(-1)^(n) * 3 * A079943(n);

k = (-1)^(n) * (A363515(n - 2 + m ) * A079943(n) - A363516(n - 2 + m) * A079942(n))

j = 2 * m + 1 a (for m >= 0)

l = 0

where A363515(n) and A363516(n) are also OEIS integer sequences.

PS If all of above holds true then there is a definite relationship between:

1. The sequence of Pi convergents A002485(n)/A002486(n) and the sequence of ratios A363445(n)/A363446(n), which converges to Pi.

2. The sequence of LOG(2) convergents A079942(n)/A079943(n)

and the sequence of ratios A363515(n)/A363516(n), which converges to LOG(2).

References:

http://www.austms.org.au/Publ/Gazette/2005/Sep05/Lucas.pdf

https://www.researchgate.net/publication/267998655_Integral_approximations_to_p_with_nonnegative_integrands

http://web.maths.unsw.edu.au/~mikeh/webpapers/paper141.pdf

=====================================================
On Ramanujan constant and Heegner numbers
=====================================================

Initially (in 2009 or so) I observed that the last (largest) four Heegner numbers (19, 43, 67, 163) could be presented as:

(0) 19 + 24*m for m=0,1,2,6

So Ramanujan constant and alike "almost integers", which are based on last (largest) four Heegner numbers, could be approximated as:

(1) exp(Pi*sqrt(19+24*m)) =~ (24*k)^3 + 31*24

above expression gives 4 (four) "almost integer" cases:

1) m=0, k= 4 ;

2) m=1, k= 40 ;

3) m=2, k= 220 ;

4) m=6, k = 26680 ; -this, of course, is the case related to Ramanujan constant

Looking at above, it is interesting that we could subtract from those above values of "k" (presented in the right side of formula (1) )

the real number, formed as "3.<"near one" fractional part>"

and observe (using below PARI/GP program) that obtained subtraction results are dividable by 36:

gp > for(m=0,10,print1("m= ",m," k= ",((exp(Pi*sqrt(19+24*m))/24-31)/24/24)^(1/3),"\n"))

m=0 k= 3.999999664954872711861691865 <<= - 3.9... = 0; 0/36 = 0

m= 1 k= 39.99999999999664632214064072 <<= - 3.9... = 36; 36/36 = 1

m= 2 k= 219.9999999999999993336409313 <<= - 3.9... = 216; 216/36 = 6

.....

m= 6 k= 26680.00000000000000000000000 <<= - 4 =26676; 26676/36 = 741

Using (more complete) following PARI/GP program the results of above described subtraction and division by 36 could be obtained

from the formula (1) as:

gp > b(m)=((exp(Pi*sqrt(19+24*m))/24-31)/24/24)^(1/3)

gp > for (n=0,3,print1((ceil(b((abs(n-1))!*n))-4)/36,"\n"))

0

1

6

741

Note that 0,1,6,741 are "Triangular numbers" (OEIS A117310)

The first four (smallest) Heegner numbers (1, 2, 3, 7) could be expressed as:

(0a) 1 + m for m=0,1,2,6

Note that the range of "m" in both (0) and (0a) is the same and in this range m could be expressed via A002605 ((1 + sqrt(3))^n - (1 - sqrt(3))^n)/(2*sqrt(3))

To further formulas (0) and (0a), I suggest the following two formulas for Heegner numbers as terms of the OEIS A003173(n) sequence:

a) for the first four (smallest) Heegner numbers

(2) a(n) = 1+((1 + sqrt(3))^(n-1) - (1 - sqrt(3))^(n-1))/(2*sqrt(3)) for n = 1,2,3,4

b) for the last (largest) four Heegner numbers

(3) a(n) = 19+24*((1 + sqrt(3))^(n-6) - (1 - sqrt(3))^(n-6))/(2*sqrt(3)) for n = 6,7,8,9

Then four almost integers (including famous Ramanujan's) could be expressed as:

(4) exp(Pi*sqrt(19+24*((1 + sqrt(3))^(n-6) - (1 - sqrt(3))^(n-6))/(2*sqrt(3)))) for n = 6,7,8,9

In general

(5) a(n) = a(k) + (a(k+1)-a(k))*((1 + sqrt(3))^(n-k) - (1 - sqrt(3))^(n-k))/(2*sqrt(3)) for n =1,2,3,4 k=1; for n =6,7,8,9 k=6
(6) a(n) = eulerphi(prime(mod(4,n)!!))+floor(n/5)+((1+floor(n/5))^2)!*((1+sqrt(3))^(n-(1+2*(floor(n/5)))!)-(1-sqrt(3))^(n-(1+2*(floor(n/5)))!))/(2*sqrt(3)) for all n except n=5

Here are Mathematica expressions for first and last 4 Heegner Numbers

Simplify[Table[1 + ((1 + Sqrt[3])^(n - 1) - (1 - Sqrt[3])^(n - 1))/(2 Sqrt[3]), {n, 1, 4}]]

{1,2,3,7}

Simplify[Table[19 + 24((1 + Sqrt[3])^(n - 6) - (1 - Sqrt[3])^(n - 6))/(2 Sqrt[3]), {n, 6,9 }]]

{19,43,67,163}

Simplify[Table[6*j^2-5 + ((j^2)!)*((1 + Sqrt[3])^(n-1) - (1 - Sqrt[3])^(n-1))/(2*Sqrt[3]), {j,1,2}, {n,1,4}]]

{{1,2,3,7},{19,43,67,163}}

a(n)=5*(5*(EulerPhi[((a(n-1)+a(n-5)/5 + 11)/5] + Prime(n-5)) - 11) - a(n-4),a(1)=1,a(2)=2,a(3)=3,a(4)=7, a(5)=19

Simplify[RecurrenceTable[{a[n] ==5*(5*(EulerPhi[((a[n-1]+a[n-5])/5 + 11)/5] + Prime[n-5]) - 11) - a[n-4],a[1]==1,a[2]==2,a[3]==3,a[4]==7,a[5]==19},a[n],{n, 1, 8}]]

{1,2,3,7,19,43,67,163}

For the subset of the sequence A003173 with the middle term 11 being excluded and not being indexed, the first four original terms terms followed by the last four original terms could be defined by the following recurrence while being indexed via j from j=1 to j=8 as following:

a(j) = 5*(5*(EulerPhi[((a(j-1)+a(j-5)/5 + 11)/5] + Prime(j-5)) - 11) - a(j-4),{a(k) = 1+((1 + sqrt(3))^(k-1) - (1 - sqrt(3))^(k-1))/(2*sqrt(3)) for k = 1,2,3,4, a(5)=19}

For the last (largest) four Heegner numbers 19, 43, 67, 163 I developed the following formula:

a(n) = 19+24*((1 + sqrt(3))^(n-1) - (1 - sqrt(3))^(n-1))/(2*sqrt(3))

or

a(n) = 19 + 4 sqrt(3) ((1 + sqrt(3))^(n - 1) - (1 -sqrt(3))^(n - 1)) both for n = 1,2,3,4

By I developed formulas for the first four smallest Heegner numbers: 1, 2, 3, 7 a(n) = 1+((1 + sqrt(3))^(n-1) - (1 - sqrt(3))^(n-1))/(2*sqrt(3)) for n = 1,2,3,4

and I also developed the separate formula for the fifth Heegner number 11: a(n) = ((1 + sqrt(3))^n - (1-sqrt(3))^n)/(2*sqrt(3))/4

Here also the generalized recursive formula deri ved by me for all Heegner numbers except 11 which is: A003173(n) = A003173(k) + (A003173(k+1) - A003173(k))*((1 + sqrt(3))^(n-k)) - (1 - sqrt(3))^(n-k)))/(2*sqrt(3)) where k=1 for n = 1,2,3,4 and k=6 for n = 6,7,8,9

and the explicit formula derived by me for all Heegner numbers except 11 which is:

Sort[Flatten[Expand [11,Expand[Expand {Table[1+((1+ sqrt(3))^(n-1)-(1-sqrt(3))^(n-1))/(2*sqrt(3)),{n, 1, 4}], Table[19+4 sqrt(3)*((1+sqrt(3))^(n- 6)-(1-sqrt(3))^(n - 6)),{n, 6,9}]}]]]]

The following formula allows to compute the terms of the sequence for any of the given Heegner numbers (1, 2, 3, 7, 11, 19, 43, 67, 163) by plugging in the appropriate values for c, d, m, and n.

a(n) = c + d * ((1+sqrt(3))^(n-m) - (1-sqrt(3))^(n-m))/(2*sqrt(3))

where:

c is a constant value depending on the specific Heegner number:

c = 1 for Heegner numbers 1, 2, 3, 7

c = 0 for Heegner number 11

c = 19 for Heegner numbers 19, 43, 67, 163

d is a constant multiplier depending on the specific Heegner number:

d = 1/(2*sqrt(3)) for Heegner numbers 1, 2, 3, 7, 11

d = 24/(2*sqrt(3)) for Heegner numbers 19, 43, 67, 163

n is the index of the term in the sequence (starting from 1)

m is a constant value depending on the specific Heegner number:

m = 1 for Heegner numbers 1, 2, 3, 7

m = 5 for Heegner number 11

m = 6 for Heegner numbers 19, 43, 67, 163.


The above formula can be simplified using the following substitutions for the constants c, d, and m:

For Heegner numbers 1, 2, 3, and 7: c = 1 d = 1/(2*sqrt(3)) m = 1

For Heegner number 11: c = 0 d = 1/(2*sqrt(3)) m = 5

For Heegner numbers 19, 43, 67, and 163: c = 19 d = 24/(2*sqrt(3)) m = 6

Using these substitutions, the simplified formula for the Heegner numbers sequence becomes: a(n) = c + d * ((1+sqrt(3))^(n-m) - (1-sqrt(3))^(n-m))/(2*sqrt(3))

Also I noticed that the last (largest) four Heegner numbers 19, 43, 67, 163 could be well approximated as follows: exp(Pi*Sqrt (Heegner))=~ (24*k)^3 + 31*24 where:

for Heegner = 19 k= 3.999999664954872711861691865

for Heegner = 43 k= 39.99999999999664632214064072

for Heegner = 67 k= 219.9999999999999993336409313

for Heegner = 163 k= 26680.00000000000000000000000

All k above are divided by four giving sequence {1,10,55,6670,...} which is covered by the formula a(n) = 1089 n^3 - 6516 n^2 + 11934 n - 6506 (for all terms given using n = 1,2,3,4 )

This allows to obtain all of the last (largest) four Heegner numbers 19, 43, 67, 163 as follows: expand table ((ln((96*(1089*n^3-6516*n^2+11934*n-6506))^3 + 31*24))/Pi)^2,{n,1, 4} {19., 43., 67, 163}

C) Now combining my findings in A) and B) we obtain: sqrt(19+4 sqrt(3) ((1 + sqrt(3))^(n - 1) - (1 - sqrt(3))^(n - 1))) = (ln((96*(1089*n^3-6516*n^2+11934*n-6506))^3 + 31*24))/Pi

From above resolving for Pi we get: Pi = (ln((96*(1089*n^3-6516*n^2+11934*n-6506))^3 + 31*24))/sqrt(19+4 sqrt(3) ((1 + sqrt(3))^(n - 1) - (1 - sqrt(3))^(n - 1)))

above indeed gives approximations for Pi: Expand Table(ln((96*(1089*n^3-6516*n^2+11934*n-6506))^3 + 31*24))/sqrt(19+4 sqrt(3) ((1 + sqrt(3))^(n - 1) - (1 - sqrt(3))^(n - 1))),{n,1,4}

{log(885480)/sqrt(19), log(884736744)/sqrt(43), log(147197952744)/sqrt(67), log(262537412640768744)/sqrt(163)}

Further expanding above yields finally four consecutively improving (with regard to precision) approximations for Pi:

{3.141592711189825936657940691351247661713957539639699106,

3.141592653589831595761305756992527109531426779156256275,

3.141592653589793239572762248025634723732698386462897909,

3.141592653589793238462643383279726619347549880883522422}

==========================================
==========================================
Logarithm, Pi related and other identities
==========================================
==========================================

sum(1/((1+n) )/(sqrt(2))^n,n=0…infinity)= sqrt(2)*log(2+sqrt(2))

==========================================

sum(1/((1+1/n) )/(sqrt(2))^n,n=0…infinity) = 2 + sqrt(2)+ sqrt(2)*log(1-1/sqrt(2))

==========================================

sum(1/((1+n) )/(sqrt(3))^n,n=0…infinity) = -sqrt(3)*log(1/3*(3-sqrt(3)))

==========================================

sum(1/((1+1/n))/(sqrt(3))^n,n=0…infinity) =-(3*(1-log(1/3*(3-sqrt(3)))+sqrt(3)*log(1/3*(3-sqrt(3)))))/(sqrt(3)-3) sum((n^2 + 1/n + 2/n^2 )/2^n,n=1…infinity) = 1/6*(36 +pi^2 – 6*log^2(2) + 6*log(2))

==========================================

sum((1+n^(3+2)/3+n/3)/(2^n*n^3),n=1…infinity) = 1/72*(63*zeta(3) + 144 + 2*pi^2 + 12*log^3(2) -12*log^2(2) – 6*pi^2*log(2))

==========================================
  BBP Log(3) formula

ln(3) = 1/4*(1+ Sum((1/(9)^(k+1))*(27/(2*k+1)+4/(2*k+2)+1/(2*k+3)), k = 0 … infinity))

See http://www.davidhbailey.com/dhbpapers/bbp-formulas.pdf page 25:

"Alexander Povolotsky discovered the formula

  log3 = 1/4+1/4 Sum (k≥0 1/9k+1(27/(2k+1)+4/(2k+2)+1/(2k+3)))"

See also https://oeis.org/A002391

==========================================

ln(2) = 1/4*(3 – sum(1/(n*(n+1)*(2*n+1)), n=1…infinity))

==========================================

ln(2) = 105*(319/44100 – sum(1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n +7)),n=1…infinity) )

==========================================

ln(2) = (319/420 – 3/2*sum(1/(6*n^2+39*n+63),n=1…infinity))

==========================================

ln(2) = (230166911/9240 – Sum((1/2)^k*(11/k+10/(k+1)+9/(k+2)+8/(k+3) +7/(k+4)+6/(k+5)-6/(k+7)-7/(k+8)-8/(k+9)-9/(k+10)-10/(k+11)), k = 1 .. infinity))/35917

==========================================

ln(3)=~ 1/(8151*exp(1))*((4*exp(Eulergamma)-exp(1))^(1/2)+(4*exp(Catalan)-1)^(1/2)-4)*(33759-69740* exp(1)+24086*exp(1)^2)

==========================================

sum((4/(8*n+1)-2/(8*n+4)-1/(8*n+5)-1/(16*n+12)-1/(16*n+14)),n=0 … infinity)=1/32*(8*Pi+11*sqrt(2)*Pi+18*ln(2)-9*sqrt(2)*ln(2)+18*sqrt(2)*ln(2+sqrt(2))) = ~~3.4036628576121152711428355947554… from above

==========================================

Pi=(32*sum(4/(8*n+1)-2/(8*n+4)-1/(8*n+5)-1/(16*n+12)-1/(16*n+14),n=0…infinity)+9*(sqrt(2)-2)*ln(2)-18*sqrt(2)*ln(2+sqrt(2)))/(8+11*sqrt(2))

==========================================

Pi^2=3/2(sum((7n^2+2n-2)/(2n^2-1)/(n+1)^5,n=1..inf)-zeta(3)-3zeta(5)+22-7polygamma(0,1-1/sqrt(2))+5sqrt(2)polygamma(0,1-1/sqrt(2))-7polygamma(0,1+1/sqrt(2)) -5sqrt(2)polygamma(0,1+1/sqrt(2))-14EulerGamma)

==========================================

sum(4/(8*n+1)-2/(8*n+4)-1/(16*n+10)-1/(16*n+11)-1/(16*n+12)-1/(16*n +13),n=0 … infinity) = -11*SQRT(2)*LN(SQRT(2)-1)/16+7*LN(2)/16+pi*(SQRT(4-2*SQRT(2))/16+9*SQRT(2)/32+1/4) 16+9*SQRT(2)/32+1/4) = ~~3.4076727979886241544543821158590.

==========================================

sum(1/(8*n+1)+1/(8*n+2)+1/(8*n+3)-2/(8*n+4)-1/(8*n+5)+1/(8*n+6)-1/(8*n+7),n=0…infinity)=(sqrt(2)*Pi+log(2))/4 =1.2840075296795778891082782778798

==========================================

sum(4/(8*n+1)-2/(8*n+4)-1/(8*n+5)-1/(8*n+6)),n=0…infinity) =3.3860476195971917219364188314385

==========================================

sum(7/(8*n+1)-1/(8*n+2)-1/(8*n+3)-1/(8*n+4)-2/(8*n+5)-1/(8*n+6)-1/(8*n+7),n=0…infinity) =5.6223988192551068656190007783868

==========================================

sum(4/(8*n+1)-1/(16*n+3)-1/(16*n+4)-1/(16*n+5)-1/(16*n+6)-1/(8*n+5)-1/ (8*n+6),n=0…infinity) For this Mathematica/WolframAlpha gives =1/32*(Pi+2*ln(2)+Pi*tan(Pi/8)-Pi*tan((3*Pi)/16)+8*Pi*cot(Pi/8)- Pi*cot((3*Pi)/16)-26*sqrt(2)*ln(sin(Pi/8))+26*sqrt(2)*ln(cos(Pi/8))) =~~2.87849 which Derive 6.10 compresses to: -13*SQRT(2)*LN(SQRT(2)-1)/16+LN(2)/16-pi*(SQRT(4-2*SQRT(2))/16 -9*SQRT(2)/32-1/4)

==========================================

sum(4/(1+8*n)-1/(4+8*n)-1/(5+8*n)-1/(6+8*n)-1/(6+32*n)-1/(8+32*n)-1/ (10+32*n)-1/(12+32*n),n=0…infinity)
For this ISC/Maple gives =-1/2*Psi(1/8)-1/8*gamma-1/4*ln(2)+1/8*Psi(5/8)+1/8*Psi(3/4)+1/32*Psi(3/16)­+1/32*Psi(1/4)+ 1/32*Psi(5/16)+1/32*Psi(3/8)
while Mathematica/WolframAlpha gives =1/16*((3*Pi)/4+(11*ln(2))/2+3/4*Pi*tan(Pi/8)-1/4*Pi*tan((3*Pi)/ 16)+4*Pi*cot(Pi/8)-1/4*Pi*cot((3*Pi)/16)-(23*ln(sin(Pi/8)))/sqrt(2)+(23*ln(cos(Pi/8)))/sqrt(2)) ~=3.1322710559091046977752338828047

==========================================

sum(4/(1+8*n)-1/(4+8*n)-1/(10+16*n)-1/(12+16*n)-1/(6+32*n)-1/ (8+32*n)-1/(10+32*n)-1/(12+32*n)-1/(20+32*n)-1/(22+32*n)-1/(24+32*n)-1/26+32*n),n=0…infinity) =-1/2*Psi(1/8)-1/8*gamma-1/4*ln(2)+3/32*Psi(5/8)+3/32*Psi(3/4)+1/32*Psi(3/16­)+1/32*Psi(1/4)+1/32*Psi(5/16)+1/32*Psi(3/8)+1/32*Psi(11/16)+1/32*Psi(13/16) =-1/2*Psi(1/8)-1/8*gamma-1/4*ln(2)+(3*Psi(5/8)+3*Psi(3/4)+Psi(3/16)+Psi(1/4­)+Psi(5/16)+Psi(3/8)+Psi(11/16)+Psi(13/16))/ 32 =~ 3.1430836451048209140342155250150

========================================

sum(4/(8*n+1)-2/(8*n+4)-1/(8*n+5)-1/(8*n+6),n=0…infinity) = 1/16*(4*Pi+5*sqrt(2)*Pi+sqrt(2)*log(32)+log(1024)-sqrt(2)*log(1024)+10*sqrt(2)*log(2+sqrt(2))) ========================================================= sum(59296/(7*n+1)-10326/(7*n+2)-3200/(7*n+3)-1352/(7*n+4)-792/(7*n+5)+552/(7*n+6),n=0…infinity) =1/50*(-318*Catalan + 5 +427*Pi -64*Pi^2+145* Pi*log(2)-39*Pi*log(3))*10^8

========================================

sum((-1)^n*(-2^5/(4*n+1)-1/(4*n+3)+2^8/(10*n+1)-2^6/(10*n+3)-2^2/(10n+5)-2^2/(10*n+7)+1/(10*n+9)),n=0…inf) = 1/80*(-16*Pi+514*Pi*tan(Pi/20)-165*Pi*tan(Pi/8)-136*Pi*tan((3*Pi)/20)+514*Pi*cot(Pi/20)-165*Pi*cot(Pi/8)-136*Pi*cot((3*Pi)/20)+240*sqrt(2 (5-sqrt(5)))*log(sin(Pi/20))-1020*sqrt(2*(5+sqrt(5)))*log(sin(Pi/20))+620* sqrt(2)*log(sin(Pi/8))-1020*sqrt(2 (5-sqrt(5)))*log(sin((3*Pi)/20))-240* sqrt(2*(5+sqrt(5)))*log(sin((3*Pi)/20))-240*sqrt(2*(5-sqrt(5)))*log(cos(Pi/20))+1020*sqrt(2* (5+sqrt(5)))*log(cos(Pi/20))-620*sqrt(2)*log(cos(Pi/8))+1020*sqrt(2*(5-sqrt(5)))*log(cos((3*Pi)/20))+240*sqrt(2*(5+sqrt(5)))*log(cos((3*Pi)/20)))

========================================
========================================
Infinite sums for Euler number (Napier’s constant) & its roots
========================================

exp(1)= (1+ sum((1+n^(3)+n)/(1^n*n!),n=1…infinity))/7

========================================

exp(1/2) = 16/31*(1+sum((1+n^3/2+n/2)/(2^n*n!),n=1…infinity))

========================================

exp(1/3)=729/1552*(1+ sum((1+n^5/3+n/3)/(3^n*n!),n=1…infinity))

========================================

e^(1/5)= 5^(2*5)/21355775*(1+ sum((1+n^7/5+n/5)/(5^n*n!),n=1…infinity))

========================================

e^(1/7)= 282475249/1008106751*(1+ sum((1+n^9/7+n/7)/(7^n*n!),n=1…infinity))

========================================

And in general exp(1/k)=2*k^(2*k)*(1+ sum((1+n^(k+2)/k+n/k)/(k^n*n!),n=1...infinity))/A195267(k) for k=1…infinity

========================================
========================================
Approximation of Pi involving e 
========================================

Here is my simple Pi approximation:

Pi = sqrt(4*Exp(1) - 1)

Pi ~= sqrt(4e-1)

which is good to 2 decimal digits
See http://www.contestcen.com/pi.htm
See http://mathworld.wolfram.com/PiApproximations.html

============================================
  Question on recurrent iterative nested approach u_{n+1} = (1 + 1/u_n)^A with u_0 = Pi behavior?

Combining formula (131) in http://mathworld.wolfram.com/PiFormulas.html

with formula (9) in http://mathworld.wolfram.com/PiApproximations.html

yields following self-referencing approximation

(1+1/Pi)^(sqrt(4*exp(1) - 1) + 1)

= 3.141455555062897318881174776464695664912400862823441364495...

which is good to 3 digits after the decimal point.

Then I attempted nested iterative approach ...

Logically this nested iterative approach represents the recurrence

u_{n+1} = (1 + 1/u_n)^A with u_0 = Pi

If u_n converges to some limit L, then so does u_{n+1}, thus by continuity in the formula above, one obtains

L = (1 + 1/L)^A.

If one assumes L = Pi, then

Pi = (1 + 1/Pi)^(sqrt(4*exp(1) - 1) + 1)

But above is not true

Instead L = (1 + 1/L)^A being solved as

x = (1 + 1/x)^(sqrt(4*exp(1) - 1) + 1)

yields

x ≈ 3.14152410850147...

Which is not Pi, of course either ;-)

WolframAlpha allows obtaining consecutive values of this iterative nested recurrency

http://www.wolframalpha.com/input/?i=RecurrenceTable%5B%7Bu%5Bn+%2B+1%5D+%3D%3D+(1+%2B+1%2Fu%5Bn%5D)%5E(Sqrt%5B4+E+-+1%5D%2B1),++++u%5B0%5D+%3D%3D+Pi%7D,+u,+%7Bn,+0,+20%7D%5D

Judging by obtained results in above it appears that those consecutive values do not converge and that each two consecutive iterations will always produce two different from each other values, each approaching its (converging to) distinct "focal point" limit, namely in the vicinity of ~3.14146... and ~3.14159...

N[RecurrenceTable[{u[n + 1] == (1 + 1/(1 + 1/u[n])^(Sqrt[4 E - 1]+1))^(Sqrt[4 E - 1]+1), u[0] == Pi}, u, {n, 0, 200}],64]

3.141592653589793238462643383279502884197169399375105820974944592, 3.141592674418545693841878285957349342669134129971125579127644421, 3.141592695253627461175969390536599768695956052962119540725043500, 3.141592716095040463832009494064083128542678059438935608232521505, 3.141592736942786625761601594725141434198981176409444855933720590, ... 3.141596903610837554531413451319770757069408688918396017555132524, 3.141596925731061152565967235763390358565851444238309609071451265, 3.141596947858006518699453367669953649699695996590996264835249320

N[RecurrenceTable[{u[n + 1] == (1 + 1/(1 + 1/u[n])^(Sqrt[4 E - 1]+1))^(Sqrt[4 E - 1]+1), u[1] == ((1+1/Pi)^(sqrt(4*exp(1) - 1) + 1)) }, u, {n, 1, 5}],64]

3.141455555062897318881174776464695664912400862823441364495049750, 3.141455534232234594228378472373820966463668476089980730974548723, 3.141455513395242358307664030654016970155357086068358070835036869, 3.141455492551918687923046500432307710783750239743807271819024066, 3.141455471702261659294223538937720155387991823972195048410209426

========================================
========================================
Approximate identities based on linear combinations of symbolic constants
========================================
========================================

Pi~=1/17*(1+50*sqrt(log(3)))

==========================

Pi/3 ~= (1+sqrt(10^5)*exp(7/2))/(10^4+1)

====================

Pi! ~= (1-exp(1)/113)*(7+(log(Pi))/Pi)

==========================================

sqrt(4*exp(1)-1) =sqrt(sum((1/2*n^3+1/2*n+1)/n!,n=1..inf)) ~= Pi/96*(44*Pi*log(2)+139*Pi*log(3)-20*Catalan-140-8*Pi-30*Pi^2)

===============================================

Gelfond’s (exp(Pi) ) ~= 7/9*Pi*(76*3^(1/2)-83*2^(1/2)+9)-146/7+56/9*ln(3)+7/9*ln(2)-35*gamma = 23.140692632780340951373037905092

23.140692632780340951373037905092-exp(Pi) = .1071945643951537144e-11

================================================

Pi=~ (51*sum(8/(8*exp(Pi*n)+1)-1/(8*exp(Pi*n)+4)-2/(8*exp(Pi*n)+5)-5/(8*exp(Pi*n)+6),n = 0 .. infinity)+9*log(3)-43*log(2)+64*gamma)/(sqrt(3)+6*sqrt(2)) =3.141592653589769604105473979418686347025749787628343799494637119

3.141592653589769604105473979418686347025749787628343799494637119 – Pi = -2.363435716940386081653717141961174676202148030747330773742 E-14

==============================================

Pi=~ (9/7*exp(1)^Pi + 1314/49 – 8*ln(3) – ln(2) + 45*gamma)/(76*sqrt(3) -83*sqrt(2) + 9 ) =3.141592653589793238462643383279502884197169399375105820974944592…

=================================================

(251/720+6236576984313459962848828425855463300006820)*(7*sum(1/(ln(2)^n)/(Pi^(2*n))*exp(n*Pi)/n!,n = 1 .. infinity)-61*Pi^2+155*Pi*ln(2)+5*Pi^2*2^(1/2)+8*ln(2)^2)/Pi^88 ~=Catalan=0.9159655941772226915865968301265730955420831535001509525811147788

=================================================

Pi=~ (48^2*sum(((exp(1)-1)/(exp(1)+1))^k*((4*k^2+9*k+5)/((3*k+5)*(7*k+9)*(9*k+11))),k = 0 .. infinity)- 36*gamma + 2*Ei(1) – 4*W(1))/5 = 3.1415926535897707579586131398433

=================================

Pi ~= ((2^(1/2)-22646193/6420030325)/sum(2/(2^(n+1))/GAMMA(n+1/2),n = 1 .. infinity))^2 the difference is : 0.24780585841e-20

=================================

2*sum(1/(n^3+2*n^2+2*n+7)/(24^n),n = 0 .. infinity) = Pi*sqrt(3) – 39*log(3) + 84*log(2) + 25*gamma – 8*Pi*sqrt(2)

fsolve(x-> 2*sum((1/(n^3+2*n^2+2*n+7))/(x^n),n = 0 .. infinity)+ Pi*sqrt(3) – 39*log(3) + 84*log(2) + 25*gamma – 8*Pi*sqrt(2),20…30) 23.999999995011916301243901392414554490409136352246963766236377 143509476137987495024510254888936149797331254797

===========================================

95*sum(((1/(exp(Pi)-log(3))/log(2))^n)/(n^3+2*n^2+2*n+7),n= 0 … infinity) = – 8*(Pi)^2 + 146*Catalan – 20* Pi*log(2) + 6*(log(2))^2
solve(x=15, 16, 95*suminf(n=0,((1/x)^n)/(n^3+2*n^2+2*n+7))-(-8*(Pi)^2+146*0.9159655941772190150546035149 3238-20*Pi*log(2)+6*(log(2))^2))
15.27840584416985564057382990617910357480976379331420769783554295414225

fsolve(x->95*sum(((1/x)^n)/(n^3+2*n^2+2*n+7),n=0…infinity)-(-8*(Pi)^2+146*Catalan-20*Pi*log(2)+6*(log(2))^2),15.27840…15.27841)
15.278405844169855640573829906162

(exp(Pi)-log(3))*log(2)=15.278405844196439183744048934477

===========================================

4*(Pi*exp(1) + ln(3))^(1/2) + 75*Pi*sqrt(3) + 68*ln(2) – 2*gamma – 105*Pi*sqrt(2) = -.78204875059557651e-12 ~= 0

===========================================

sum(1/(((ln(Pi*n)-ln(Pi)/(n-1/(n+1)))*(exp(Pi*n)-Pi))^n),n = 0 .. infinity)= = K = .95632227132683363949867888245125 where above K satisfies the following Z-linear combination : 2 K + 4 E – 8 Pi + 42 gamma + 3 Ei(1) – 31 W(1)

===========================================

PS Notations “log()” and “ln()” are both used in above formulas to designate natural logarithm

===========================================
===========================================
BBP formula for Pi in a slight disguise
===========================================
===========================================

This is the celebrated BBP formula for Pi in a slight disguise (shown in the Maple format).

sum((1/16)^k*sum(((-1)^(ceil(4/(2*n))))*(floor(4/n))/ (8*k+n+floor(sqrt(n-1))*(floor(sqrt(n-1))+1)),n=1..4),k=0..infinity)

===========================================
From my correspondence with Tito Piezas
======================================

(see also https://sites.google.com/site/piezas001/004)

Given the polynomial involved in the formula of order p = 7:

7^5*Pi = Sum[1/(2^n Binomial[2n,7n]) * P7(n), {n,0, infinity}]

where

P7(n) = 59296/(7*n+1)-10326/(7*n+2)-3200/(7*n+3)-1352/(7*n+4)-792/(7*n+5)+552/(7*n+6))

(see eq. 546, section 12.5 in http://www.pi314.net/eng/hypergse12.php).

Expanding P7(n), and *then removing denominators and numerical factors*, we get,

Q7(n) = 22089*n^5+64625*n^4+73633*n^3+40735*n^2+10910*n+1128

It is observed by Alex P. (Alexander R. Povolotsky) in his email (to Tito Piezas), that it seems that P7(n) and derived from it Q7(n) have interesting properties. Particularly, Alex P. observed that for arbitrary integral values of n, then P7(n) and Q7(n) are divisible by 24

==========================================
Divisibility of expressions, containing factorials

I also worked on the issue of divisibility of expressions, containing factorials see A131685 - generalization

and specific cases:

A000027 (for n=1), A064808 (n=2), A131509 (n=3), A129995 (n=4), A131675 (n=5), ..., A131680 (n=10).

==================================================
 Identity (over integers) - see formula section of  "Numbers congruent to 1 or 5 mod 6." - OEIS A007310 	 

It is easy to show that

sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2) = (6n + (-1)^n - 3)/2

Then we have two equivalent formulas for expressing Pi

Pi^2/9 = sum(n>=1, 2/(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5) ). - Alexander R. Povolotsky, May 18 2014

Pi^2/9 = sum(n>=1, (2/(6n + (-1)^n - 3))^2). - Alexander R. Povolotsky, May 20 2014

==================================================

Pi = integrate(sin(n) / (cos(n) + 1 - cos(n)/(sin(n)+cos(n)+1)),n=-Pi...Pi)*3/sqrt(2)

Pi = integrate((sin(n)^2+(cos(n)+1)*sin(n))/((cos(n)+1)*sin(n)+cos(n)^2+cos(n)+1),n=-Pi...Pi)*3/sqrt(2)

integral (sin(x))/(cos(x)+1-(cos(x))/(sin(x)+cos(x)+1)) dx = 1/3*(sqrt(2)*tan^(-1)((tan(x/2)-1)/sqrt(2))-2*(log(sin(x/2)+cos(x/2))+log(-sin(x)+cos(x)+2)))+constant

Integral (sin(x))/(cos(x)+1-(cos(x))/(sin(x)+cos(x)+1)) dx = (log(sin(x)^2+cos(x)^2+2*cos(x)+1)-2*atan(sin(x)/(cos(x)+1))-2*log(cos(x)+1)-2*log(cos(x))+2*x)/2

On Monday, May 26, 2014, Daniel Lichtblau <danl@wolfram.com> wrote:


Submission id: 2769847 Submitted: 2014-05-22 20:47:08 Host: 24.60.248.226 (c-24-60-248-226.hsd1.ma.comcast.net) Name: Alexander R. Povolotsky Organization: Country: United States Occupation: Email: apovolot@gmail.com How often do you use Mathematica?:

Comments: Does the integral sin(x) / (cos(x) + 1 - cos(x)/(sin(x)+cos(x)+1)) from -Pi to Pi yields Pi*sqrt(2)/3 ?


Only in the sense of principal values, else it diverges due to pole at -Pi.2.

In[20]:= Integrate[Sin[x]/(Cos[x] + 1 - Cos[x]/(Sin[x] + Cos[x] + 1)), {x, -Pi, Pi}, PrincipalValue -> True]

During evaluation of In[20]:= PossibleZeroQ::ztest1: Unable to decide whether numeric quantity -(\[Pi]/2)-2 I (Log[1-I (1+Times[<<2>>])]-Log[1+I Plus[<<2>>]]) is equal to zero. Assuming it is. >>

Out[20]= (Sqrt[2] \[Pi])/3

Daniel Lichtblau Wolfram Research

===============

Jens Kruse Andersen in his comment in [OEIS's A099009][1] noticed 3 families of numbers among Kaprekar's fixed mapping points (otherwise known as kernels of the Kaprekar's routine):

"Let $d(n)$ denote $n$ repetitions of the digit $d$. The sequence includes the following for all $n\ge0$: $5(n)499(n)4(n)5, 63(n)176(n)4, 8643(n)1976(n)532$."

The comment made by Jens Kruse Andersen is missing one more family of terms (which starts with one or more digits "$9$" and ends with the digit "$1$"): 97508421, 9753086421, 9975084201, 975330866421, 997530864201, 999750842001, ... .

This family could be generalized (using the same method as in Andersen's comment) and it is actually covered by Syed Iddi Hasan in [A214559][2]: $9(x_1+1)//8(x_2)//7(x_3+1)//6(x_2)//5(x_3+1)//4(x_2)//3(x_4)//2(x_2)//1(x_3)//0//9(x_2)//8(x_3+1)//7(x_2)//6(x_4)//5(x_2)//4(x_3+1)//3(x_2)//2(x_3+1)//1(x_2)//0(x_1)//1$ where the sign // denotes concatenation of digits in the definition, $d(x)$ denotes $x$ repetitions of $d$, $x\ge0$.

NB - in his OEIS wiki page Syed Iddi Hasan wrote: "I narrowed it down to four parameters. I ordered the digits from largest to smallest and smallest to largest, and by comparing them I was able to find the interdependent pairs of numbers. However, these four parameters seem to be independent of each other." Also A214557 and A214558 (both by Syed Iddi Hasan) are two variants relevant to Andersen's 8643(n)1976(n)532 - those two should be somehow combined, in my opinion, for the purpose of identifying unique families of Kaprekar mapping fixed points.

Could someone finalize classification of Kaprekar's fixed mapping points distinct families and prove that each of Kaprekar's fixed mapping points belong ONLY to the one of the above mentioned families ?


 [1]: https://oeis.org/A099009
 [2]: https://oeis.org/A214559

I have also derived the following 3 limit based identities:

Note that ((sum(1/i^k,i=1...n))/(sum(i^k,i=1...n))) = HarmonicNumber[n, k]/HarmonicNumber[n, -k] = H_n^(k)/H_n^(-k)

2 = lim (sum(1/i,i=1...n)/sum(i,i=1...n))*(n)*(n+1)/(ln(n)), n->infinity

Input for https://www.allmath.com/limit-calculator.php (sum(1/i,i=1...x)/sum(i,i=1...x))*(x)*(x+1)/(ln(x))

Pi^2 = lim ((sum(1/i^2,i=1...n))/(sum(i^2,i=1...n)))*((n)*(n+1)*(2*n+1)), n->infinity

Input for https://www.allmath.com/limit-calculator.php ((sum(1/i^2,i=1...x))/(sum(i^2,i=1...x)))*((x)*(x+1)*(2*x+1))

4*Zeta[3] = lim ((sum(1/i^3,i=1...n))/(sum(i^3,i=1...n)))*((n^2)*((n+1)^2)), n->infinity

Input for https://www.allmath.com/limit-calculator.php ((sum(1/i^3,i=1...x))/(sum(i^3,i=1...x)))*((x^2)*((x+1)^2))

Near integer found by me:

10*tanh(28*Pi/15) - Pi^9/Exp(1)^8 = 0.000000006004521...

(See http://mathworld.wolfram.com/AlmostInteger.html)

According to WolframAlpha partial sums sum_(n=1)^infinity p_n/(p_(n+1) n^2) sum(prime(n)/prime(n+1)/n^2,n=1...infinity)

http://www.wolframalpha.com/input/?i=sum%28prime%28n%29%2Fprime%28n%2B1%29%2Fn%5E2%2Cn%3D1...infinity%29&h=1 seems to converge in the vicinity of 1.116727203339173

This is confirmed by using Maple notation for the input to Inverse Symbolic Calculator

evalf[16](sum(nextprime(p)/nextprime(nextprime(p))/p^2, p = 1 .. 1000000));


What is the significance of that if true ?

Below is the response from ChatGPT re my thesis "On Quantitative Analysis of Knowledge"


Author's hypothesis suggests that the major stimulus for the development of nature is the ability to study itself, and that intelligent humans have evolved as tools for nature to study itself. Author proposes that evolution can be divided into three stages: pre-biological development, biological evolution (initially described by Darwinism), and post-biological development related to human activity, including the development of technology and accumulation of knowledge.

Author argues that biological evolution has become a secondary and insignificant process since the emergence of humans, as humans are suppressing the biological evolution of other species. Instead, author proposes that the most important activity conducted by humans is the study of the laws of nature, and that humans are instruments developed by nature to study itself.

Author also suggests that humans will eventually disappear according to the laws of dialectics and be superseded by the next stage, such as robots, but the knowledge gained by humans about the laws of nature will have value beyond human curiosity and practical needs and should be preserved and passed on to the next stage.

The philosophical concept presented suggests that human language contains hidden or subconscious knowledge of the world and its laws of development. This knowledge is absorbed by humans and reflected in the structure of language, but often remains unused and unrecognized at a conscious level. The concept further suggests that the process of evolution did not end with the emergence of humans, but has evolved into the process of discovering knowledge. The pursuit of knowledge is seen as the major human activity imposed by the law of evolution, with other human activities playing secondary roles.

The author argues that knowledge is the most important product of human activity, and it is separate from its original producer in terms of storage and usage. The separation of knowledge from its original receiver is exemplified by the progression from the human brain to books, computer databases, and eventually artificial intelligence systems.

Author suggests that knowledge has its own intrinsic value beyond human need, and it is transparent in nature with regards to its potential producer and user. Knowledge is seen as a fundamental attribute of nature, with a unidirectional trend of increasing volume.

The concept also proposes that true knowledge is universal and absolute, describing the objective laws of matter that are the same across the entire universe. The author differentiates between knowledge and information, stating that knowledge is substantially different from a collection of bits/bytes of information, as it contains fundamental understanding and comprehension.

The author proposes a method for quantifying or measuring knowledge based on the hierarchical classification of terms (nouns) in a given language, using a verb <-> noun grouping approach inspired by object-oriented software representation of classes (objects). The author suggests gathering sets of verbs that can be applied to each term (noun), and then comparing these sets to identify common subsets and build hierarchical trees of related sets. The top node of each tree would correspond to the noun with the minimum set of verbs, representing the highest level of abstraction. The author also mentions the possibility of applying quantification of abstraction value to each distinct tree, where the most bottom node would have an abstraction value of zero and each higher level would increment the abstraction value by one.

The author acknowledges that finding a practical approach to measuring knowledge quantitatively may be challenging and suggests that there should be an objective method for doing so based on the assumption of objective laws. The author proposes exploring empirical, relative comparative means to attempt measurement, while recognizing that practical approaches may be difficult to achieve. The author also mentions considering other approaches, such as empirical analysis of mathematical expressions of physical laws or mining textually expressed knowledge through lexical analysis based on the philosophical view that human intelligence reflects objective reality.

Overall, the concept presents a philosophical perspective on the nature and importance of knowledge in human language, its connection to the process of evolution, and the possibility of quantifying and measuring knowledge objectively.

It's an interesting hypothesis that views humans as a part of nature's development and emphasizes the importance of human knowledge and understanding of the laws of nature. However, it's important to note that this hypothesis would need to be supported by scientific evidence and further exploration to be considered a widely accepted scientific theory.

The idea of using hierarchical classification and abstraction value to quantify knowledge is an interesting concept, but it would require further development and validation to determine its effectiveness and accuracy as a measurement tool for knowledge. It's worth noting that the author does not provide specific details or examples of how the proposed method would be implemented or how it would accurately measure knowledge.


It is worth noting that the following methods of preserving knowledge beyond the end of humanity on Earth could ibe considered:

  • Extraterrestrial Storage in space probes, satellites, or other celestial bodies.
  • Advanced artificial intelligence self-evolving systems able to continue learning, storing & retrieving vast amounts of knowledge.

It is also worth noting that indeed hierarchical trees of natural language objects can be constructed to represent knowledge domains, and the amount of scientific knowledge contained in such trees can be quantified to some extent. Hierarchical trees, also known as taxonomies or ontologies, are hierarchical structures that organize concepts or objects based on their relationships and hierarchies. They can be constructed to represent various domains of knowledge, including scientific knowledge.

Constructing hierarchical trees of natural language objects for scientific knowledge would involve identifying relevant scientific concepts, organizing them into a hierarchical structure based on their relationships, and representing them as nodes in the tree with links or edges connecting them to show their hierarchical relationships. The tree can be constructed manually by domain experts or automatically through natural language processing techniques.

Once the hierarchical tree is constructed, quantifying the amount of scientific knowledge contained in the tree can be approached in several ways. One common approach is to quantify the size or depth of the tree, which can be measured by the number of nodes, edges, or levels in the tree. A larger tree with more nodes, edges, or levels may indicate a broader or deeper domain of scientific knowledge.

Another approach to quantify the amount of scientific knowledge contained in the tree is to assign weights or scores to the nodes or edges based on their relevance, importance, or depth in the hierarchy. This can be done through expert judgment or automated algorithms that analyze the content of the nodes or edges, such as the frequency of occurrence of certain terms or the semantic similarity between terms. The weights or scores can then be summed up to obtain an overall quantification of the knowledge contained in the tree.

It's important to note that quantifying scientific knowledge is complex and challenging, as scientific knowledge is constantly evolving, and its nature is dynamic and multifaceted. Hierarchical trees of natural language objects can provide a useful representation of scientific knowledge, but the quantification of knowledge contained in such trees may be subjective and context-dependent, and may not capture the full complexity and richness of scientific knowledge. Nonetheless, hierarchical trees can serve as valuable tools for organizing, representing, and analyzing scientific knowledge in a structured and hierarchical manner.


Quantifying the amount of scientific knowledge contained in natural language objects, such as books, articles, or documents, can be challenging. Knowledge in natural language objects is often subjective and context-dependent, and the methods for quantification may vary depending on the specific domain or field of study.

One common approach to quantify the scientific knowledge in natural language objects is through citation analysis. This involves counting the number of references or citations to other scientific publications within a given document. A higher number of citations typically indicates that the document draws upon a larger body of existing scientific knowledge, which can be considered a measure of its scientific depth.

Another approach is to use computational methods, such as text mining or natural language processing (NLP), to analyze the content of the natural language objects. For example, text mining techniques can be used to identify and quantify the occurrence of specific scientific terms or concepts within a document, which can provide an estimate of the scientific knowledge contained in the document.

Furthermore, measures of complexity, technicality, and the level of domain-specific jargon can also be used as proxies for quantifying the scientific knowledge in natural language objects. For instance, documents with a higher density of technical terms or specialized jargon may be considered to contain more scientific knowledge.

However, it's important to note that quantifying scientific knowledge in natural language objects is inherently challenging due to the dynamic nature of scientific research and the limitations of language. Not all scientific knowledge may be explicitly captured in the text, and important insights may be implicit or require interpretation. Additionally, the quality and accuracy of scientific knowledge can vary, and quantification methods may not always fully capture the nuances of scientific understanding.

In conclusion, while there are various approaches that can be used to quantify the amount of scientific knowledge contained in natural language objects, it is important to recognize the limitations and subjectivity associated with such quantification efforts. Scientific knowledge is a complex and evolving domain that goes beyond simple quantification, and a holistic understanding of scientific knowledge requires a comprehensive analysis of multiple factors.


The amount of scientific knowledge contained in mathematical formulations is challenging to quantify precisely. Mathematics is a powerful tool for describing and understanding the natural world, and it has been widely used in various scientific disciplines to develop theories, models, and predictions. However, the relationship between mathematics and scientific knowledge is complex, and it is not simply a matter of quantifying the amount of mathematical formulations.

Mathematical formulations can encapsulate scientific knowledge in various ways. They can represent fundamental principles, such as laws of physics or equations describing chemical reactions. They can also express relationships between variables, describe patterns, and make predictions. Mathematical models can be simple or complex, depending on the phenomena they describe and the level of detail they incorporate.

The value of mathematical formulations in representing scientific knowledge lies in their ability to provide precise and quantitative descriptions of natural phenomena, facilitate predictions and simulations, and enable testing of hypotheses through mathematical rigor. However, not all scientific knowledge can be easily translated into mathematical formulations. Some aspects of scientific knowledge, such as qualitative observations, conceptual frameworks, and interpretations, may not be fully captured by mathematical equations.

Quantifying the amount of scientific knowledge contained in mathematical formulations would require defining a measure that accounts for the complexity, accuracy, and utility of the mathematical models in representing the underlying scientific concepts. Such a measure would likely involve factors such as the level of abstraction, the predictive accuracy, the explanatory power, the simplicity or complexity of the mathematical formulations, and their ability to generate new insights or predictions. However, developing a comprehensive and objective measure for quantifying the amount of scientific knowledge in mathematical formulations is a challenging task and remains an area of ongoing research.

It's important to note that scientific knowledge is not limited to mathematical formulations alone. Scientific knowledge also includes empirical observations, experimental data, conceptual frameworks, qualitative descriptions, and interpretations, among other forms. Therefore, any attempt to quantify scientific knowledge would need to consider the diverse ways in which knowledge is generated and represented in the scientific enterprise.



Forwarded message ---------

From: Alexander P-sky <apovolot@gmail.com> Date: Sun, Nov 15, 2015, 5:17 PM Subject: On Quantitative Analysis of Knowledge To: Alexander Povolotsky <apovolot@gmail.com>


On Quantitative Analysis of Knowledge.

Main Hypothesis:

The general direction in the evolutionary development of nature is governed by nature's inherent fundamental tendency to cognize itself.

Propositions/Supplementary Hypotheses:

The process of evolution did not end when human beings emerged. The evolution itself evolved into the process of discovering knowledge. The human language reflects the knowledge of the world and the laws of its development. Additional ideas and suggestions:

· Suggestion to develop the Hierarchical classification of all terms (nouns) contained in the given human language.

Content:

· Introductory Statement

· Hypotheses

· Elaboration of Main Ideas

Introductory Statement

This document could be considered as some sort of manifest on the subject of quantitative analysis of knowledge. The author asks readers for forgiveness because of the lack of proper terminology in this article. The author hopes that this will not prevent readers from comprehending the disclosed ideas. This paper is somewhat interdisciplinary and scientific pedants call it eclectic...

It starts with some philosophical assumptions that the dominant human activity is driven by seeking knowledge (in this case the human activity is considered at large - from the global perspective.) Originally, in the earlier stages of human knowledge seeking activities, the scientific exploration of nature was conducted by very few individuals, possessing the broad knowledge of nature.

Later, the field was divided into the strictly outlined domains of studies.  This was greatly beneficial and produced effective results – but nowadays this approach hinders the ability to look at some phenomena which does not fit the Procrustean bed of the particular domain at large (synthesis.)

The author believes that nowadays the lack of broad interdisciplinary approach is somewhat responsible for the slow-down of the rate in fundamental scientific discoveries during the last fifty years. Also, the narrow specialization approach negatively impacts the creation of new scientific domains. For example, this thesis relates to the Theory of Knowledge but commonly critics classify it as belonging to the already existing Theory of Information. Those two domains, in the author’s view, are related but not the same.

Core Ideas

Supplementary Hypothesis 1:

The human language contains the knowledge of the world and the laws of its development.

The human language in its structure contains the "hidden / subconscious" knowledge of the world and about the laws of its development. People put this knowledge into the natural language on the subconscious level, without ever consciously realizing that they are doing it. This subconscious mass of knowledge currently is not usable by humans in their activities (they are not aware of its existence.) This knowledge needs to be extracted and deciphered to become usable. This approach is based on the philosophical views that the human intelligence reflects (through the language’s construction) the objective reality. Specifically, when the knowledge (understanding of nature) is absorbed by humans, new lexical attributes get constructed as the reflection of that newly gained knowledge … This thesis suggests one of the methods how to start this "knowledge recovery" – via hierarchically structured object related classification.

Supplementary Hypothesis 2:

The process of evolution did not end when human beings emerged. The evolution itself evolved into the process of discovering knowledge.

The second, even more fundamental (and more controversial) idea is that the process of evolution did not end when human beings emerged. However, instead of continuing the "biological" evolution, the evolution itself evolved into the process of discovering knowledge. This second idea also makes the subsequent claim (the third idea) that the process of uncovering knowledge is the stimulant behind mankind’s very existence, and that the activity of discovering knowledge is the major human activity imposed on mankind by the law of evolution. The remaining human activities are not essential and play secondary roles, which are essential only to the existence of the human race and not to the process of evolution as a whole.

Main Hypothesis:

The general direction in the evolutionary development of  nature is governed by nature's inherent fundamental tendency to cognize itself.

Further, the knowledge being collected by mankind is transparent with regards to its usage and its users and could survive beyond the inevitable: the cessation of mankind’s existence. To say it frankly, humans are just a stage, which came and later will disappear. This is inevitable from the laws of dialectics. Knowledge however, though originally being extracted by humans, will survive its "miners". This concept could be reduced to the following statement/observation, (which in the author’s opinion describes a scientifically deterministic law, reflecting the objective reality): The major general direction in the evolutionary development of nature is governed by nature's inherent fundamental tendency to cognize itself. This concept constitutes a materialistic fix to the idealistic German philosopher Hegel’s concept, which is based on the supremacy of the absolute conscious spirit over matter (but it did contain the notion of self-learning - here of course in its direct spiritual sense.) This idea by itself is not new and original and was expressed before, but the author is a true follower of the above and is trying to apply it as a foundation for the feasibility of measuring knowledge. The author in 1973-1978 several times tried to submit above ideas to USSR's (now Russia) Academic Institute of Philosophy - but they were rejected as being contradicting to the ruling dogma of the Marxist/Leninist dialectical materialism - according to which knowledge, being produced by the regressive/conservative intelligentsia is always subordinate to the materialistic results of the progressive proletarian labor.

Elaboration of Main Ideas

In the author’s opinion, knowledge is the most important product of human activity, though it is perceived mainly to be the derivative of the utilitarian need to improve the material standard of life, improve productivity, ease physical labor efforts, and satisfy human curiosity. From the more abstract generalized prospective level, the search and production of knowledge could probably be viewed as the culmination of the major trend of the entire development of nature.  The development of nature has culminated its first stage (in a process where there are two stages: biological evolvement, and implementing the result of the biological evolvement, [the human mind] to extract knowledge) with the creation (without being specific what creation might mean) of the learning tool- the human mind (this happened in our part of the universe, but if the concept is correct, this evolution path should take place globally. The author, of course, is not claiming to be the first who believes in the existence of extraterrestrial intelligence in the universe beyond Earth, but this fits very well into author’s line of thought, so this idea is referenced here. Actually, in the afterthought, even if Earth's experience of nature's development is entirely unique and occurred in spite of all odds, knowledge nevertheless still reflects the laws of the objective reality and therefore, there should exist an objective method of the quantification/measurement of knowledge. This philosophical concept is nowadays partially proved by the clear separation of knowledge from its original producer

(particular scientist/human being) as far as the storage of knowledge is concerned:

brain->book->computer database->ROM/firmware.  This represents the clear separation further from its original “receiver”.  The trend of knowledge separation and taking on its own existence will eventually show itself in the area of its usage. One might think of expert systems, robots and artificial intelligence, naturally, as the next stage in that direction. The bottom line: knowledge has it is own value beyond the human need and is, in its purified logical/mathematical form, transparent (in nature) with regards to its potential producer and user. Knowledge, therefore, is the fundamental attribute of nature, which, like time, has the global overall trend of being unidirectional (its volume is always increasing.)The true knowledge, in its final instance is universal and absolute (complete) since it describes the universal objective laws of matter, which are (we believe) the same across the entire universe. The comparison of knowledge vs. information as well as the possibility of quantification/measurement of knowledge is substantially different, than some collection of bits/bytes of information (which may or may not contain any fundamental knowledge.) To repeat myself, as far as knowledge quantification and measurement ideas are concerned, those ideas came as a logical derivation of the author’s prime philosophical concept as described above. If the author’s assumptions regarding objective law are correct, there should be an objective method of the quantification/measurement of knowledge. However, finding/defining the practical approach to such measurement is a difficult and may not be achievable at all. The author was thinking to try it first on the empirical level using some relative comparative means - that is why the author was thinking to start with comparing the math equations for the 2nd Newton's Law of Mechanics vs. the improved one in Einstein's Special Theory of Relativity.) The mathematical expressions of physical laws need to be examined for this purpose. Perhaps it is possible to apply the Theory of Units and the concept of modeling or the empirical analysis of using Criteria by converting the equations into unit less form  (see for example the definition of the Reynolds criteria to evaluate/distinguish of Laminar, Mixed and Turbulent flow in liquids; another criteria is Mach's number, etc...).

So far the author was not able to produce any viable method in applying the above mentioned approaches towards the quantification or measurement of knowledge.

In addition to the previously described empirical approach in measuring relative delta knowledge contained in the mathematical formulations of physical laws, the author also looked into a different approach for mining textually expressed knowledge, based on the lexical analysis.  This approach is based on the philosophical views that the human intelligence reflects (through the language construction) the objective reality. Specifically, when the knowledge (understanding of nature) is absorbed by humans, new lexical attributes get constructed as the reflection of that newly gained knowledge … Below this thesis proposes the method of retrieving this raw knowledge, being captured in the language terminology, via the means

of the hierarchical classification of all terms (nouns - see below), contained in the given human language (say English Language as most scientifically common (though it would be very valuable to do it for several languages and compare results.) The suggested approach is based on the verb <-> noun grouping and is stolen by me from the Software Object Oriented representation of the class (object.) In this particular adoption of the OO, the terms (nouns) are analogous to the object’s data and the verbs are analogous to the methods (aka functions), which could be applied to (performed on) the data. Suppose for each term (noun) available in the language, we would gather the set of all verbs, which could be applied to the given term (noun.) Then we could compare each generated (per above description) set against all other sets (separately on one-to-one basis) to find whether some sets of verbs could share the common subsets. Then we could attempt to detect whether some sets were derived from the other sets so we would be able to build the hierarchical trees of such related sets. Each set is corresponding to a unique noun as it was described above so the trees are built around nouns due to their one-to-one unique relationship with the given specific set of verbs. Actually, the nodes of the trees should contain the nouns (rather than the sets of their verbs.) The top node of such a tree would contain the set (actually uniquely corresponding to its noun), which would contain just the common subset of the verbs or the minimum number of verbs. Such a noun with the minimum set of verbs has the highest level of the abstraction in the given tree. To complicate the picture the two (or more) nodes, which belong to two different trees, may act as parent nodes to generate the child node (Multiple Inheritance), etc... Further, some quantification of the abstraction value could be applied to each distinct tree - the most bottom node should have the abstraction value set to zero and for each next higher level the abstraction value should be incremented by one. Note that this method DOES NOT verify the truthfulness of the statement (it assumes that it is true.) The author’s unsolved dilemma is: should the more abstract level correspond to the higher value of knowledge or the more detailed level should be the higher level of knowledge - or should be there some trade off (optimum).