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# User:Alexander R. Povolotsky

Alexander R. Povolotsky apovolot@gmail.com

Below identity is quite trivial and may even be called superficial but to me it has some beauty in it ...

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

Near integer found by me:

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

Here is my simple but beautiful Pi approximation:

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

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


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

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

Problems & Puzzles: Conjectures. Conjecture 59.

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

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

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

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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

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Base specific "equal ratio" pairs of distinct permutations

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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

${\displaystyle 9876543210/1234567890=8.0000000729...}$

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

${\displaystyle 7901234568/(0)987654312=8.0000000729...}$

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) ${\displaystyle (n^{2}(n+1)^{n}-(n+1)^{n}+1)/(-n^{2}+n(n+1)^{n}+(n+1)^{n}-n-1)}$

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)

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

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Identity for 24/Pi

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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

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  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

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

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k-fold nested sum of integer powers

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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:

and

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

comment sections of the following OEIS sequences:

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.

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Generalizing Stephen Lucas's identities for Pi and its convergents

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I conjecture that the following identity below represents a generalization of Stephen Lucas's (see references to his publications at the end of this section? experimentally obtained identities between Pi and its convergents:

${\displaystyle (-1)^{n}\cdot (\pi -{\text{A002485}}(n)/{\text{A002486}}(n))=(|i|\cdot 2^{j})^{-1}\int _{0}^{1}{\big (}x^{l}(1-x)^{m}(k+(i+k)x^{2}){\big )}/(1+x^{2})\;dx}$ (1)

In unformatted form:

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

where integer n = 3,4,5,... serves as the index for terms in OEIS A002485(n) and A002486(n),

and {i, j, k, l, m} are some integers (to be found experimentally or otherwise), which are probably some functions of n.

The "interesting" (I think) part of my generalization conjecture is that "i" is present in both:

the denominator of the coefficient in front of the integral and in the body of the integral itself.

1) For example, in cited by Lucas old known formula for 22/7

22/7 - Pi = Int(x^4*(1-x)^4*/(1+x^2),x = 0 .. 1)

${\displaystyle {\frac {22}{7}}-\pi =\int _{0}^{1}{\frac {x^{4}(1-x)^{4}}{1+x^{2}}}\,\mathrm {d} x}$

with n=3, i=-1, j=0, k=1, l=4, m=4 - with regards to my above suggested generalization.

In Maple notation

i:=-1; j:=0; k:=1; l:=4; m:=4; Int(x^l*(1-x)^m*(k+(k+i)*x^2)/((1+x^2)*(abs(i)*2^j)),x= 0...1)

yields 22/7 - Pi

2) And in Lucas's formula for 333/106

Pi - 333/106 = 1/530*Int(x^5*(1-x)^6*(197+462*x^2)/(1+x^2),x = 0 .. 1)

${\displaystyle \pi -{\frac {333}{106}}={\frac {1}{530}}\int _{0}^{1}{\frac {x^{5}(1-x)^{6}(197+462x^{2})}{1+x^{2}}}\,\mathrm {d} x}$

with n=4, i=265, j=1, k=197, l=5, m=6 -with regards to my above suggested generalization.

In Maple notation i:=265; j:=1; k:=197; l:=5; m:=6; Int(x^l*(1-x)^m*(k+(k+i)*x^2)/((1+x^2)*(abs(i)*2^j)),x= 0...1)

yields Pi - 333/106

3) In Lucas's formula for 355/113

355/113 - Pi = 1/3164*Int(x^8*(1-x)^8*(25+816*x^2)/(1+x^2),x = 0 .. 1)

${\displaystyle {\frac {355}{113}}-\pi ={\frac {1}{3164}}\int _{0}^{1}{\frac {(x^{8}(1-x)^{8}(25+816x^{2})}{(1+x^{2})}}}$

with n=5, i=791, j=2, k=25, l=8, m=8 -with regards to my above suggested generalization.

In Maple notation

i:=791; j:=2; k:=25; l:=8; m:=8; Int(x^m*(1-x)^l*(k+(k+i)*x^2)/((1+x^2)*(abs(i)*2^j)),x= 0...1)

yields 355/113 - Pi

4) In Lucas's formula for 103993/33102

Pi - 103993/33102 = 1/755216*Int(x^14*(1-x)^12*(124360+77159x^2)/(1+x^2),x = 0 .. 1)

${\displaystyle \pi -{\frac {103993}{33102}}={\frac {1}{755216}}\int _{0}^{1}{\frac {x^{14}(1-x)^{12}(124360+77159x^{2})}{1+x^{2}}}\,\mathrm {d} x}$

with n=6, i= -47201, j=4, k=124360, l=14, m=12 -with regards to my above suggested generalization.

In Maple notation

i:=-47201; j:=4; k:=124360; l:=14; m:=12; Int(x^l*(1-x)^m*(k+(k+i)*x^2)/((1+x^2)*(abs(i)*2^j)),x= 0...1)

yields Pi - 103993/331

5) In Lucas's formula for 104348/33215

104348/33215 - Pi = 1/38544*Int(x^12*(1-x)^12*(1349-1060*x^2)/(1+x^2),x = 0 .. 1)

${\displaystyle {\frac {104348}{33215}}-\pi ={\frac {1}{38544}}\int _{0}^{1}{\frac {x^{12}(1-x)^{12}(1349-1060x^{2})}{1+x^{2}}}\,\mathrm {d} x}$

with n=7, i= -2409, j=4, k=1349, l=12, m=12 - with regards to my above suggested generalization.

In Maple notation

i:=-2409; j:=4; k:=1349; l:=12; m:=12; Int(x^l*(1-x)^m*(k+(k+i)*x^2)/((1+x^2)*(abs(i)*2^j)),x= 0...1)

yields 104348/33215 - Pi

6) And it works as well for the convergent

${\displaystyle {\frac {618669248999119}{196928538206400}}}$

which is not part of A002485(n) /A002486(n) convergents sequence

${\displaystyle {\frac {618669248999119}{196928538206400}}-\pi ={\frac {1}{755216}}\int _{0}^{1}{\frac {x^{14}(1-x)^{12}(77159+124360x^{2})}{1+x^{2}}}\,\mathrm {d} x}$

with i= 47201, j=4, k=77159, l=14, m=12 -with regards to my above suggested generalization.

In Maple notation

i:=47201; j:=4; k:=77159; l:=14; m:=12;Int(x^l*(1-x)^m*(k+(k+i)*x^2)/((1+x^2)*(abs(i)*2^j)),x= 0...1)

yields 618669248999119/196928538206400 - Pi

Matt B. has provided analytical proof and improved the formula by reducing the number of parameters from 5 to 4

${\displaystyle (-1)^{n}(\pi -{\frac {p_{n}}{q_{n}}})=\int _{0}^{1}{\frac {x^{\epsilon +2m'}(1-x)^{2m'}(\alpha +\beta x^{2})}{(\alpha -\beta )2^{m'-2}(-1)^{\epsilon }(1+x^{2})}}dx}$

Below is the list of parameters in Matt B.'s formula for all cases, covered in Stephen Lucas' publications - the stuff to the right of the arrow sign is the actual Maple code, which one could copy (while in the "edit" mode) and then paste into (let say) Inverse Symbolic Calculator (it accepts Maple code) and run it there.

NB I replaced for brevity some parameter names used by Matt B: "alpha" by "a", "beta" by "b", "epsilon" by "c", "m' " by "p".

104348/33215 - Pi -> a:=1349;b:=-1060;p:=6;c:=0;Int((x^(c+2*p)*(1-x)^(2*p)*(a+b*x^2))/((a-b)*2^(p-2)*‌((-1)^(c)*(1+x^2))),x=0...1)

Pi - 103993/33102 -> a:=124360;b:=77159;p:=6;c:=2;Int((x^(c+2*p)*(1-x)^(2*p)*(a+b*x^2))/((a-b)*2^(p-2‌)*((-1)^(c)*(1+x^2))),x=0...1)

355/113 - Pi -> a:=25;b:=816;p:=4;c:=0;Int((x^(c+2*p)*(1-x)^(2*p)*(a+b*x^2))/((a-b)*2^(p-2)*((-1‌)^(c)*(1+x^2))),x=0...1)

Pi - 333/106 -> a:=197;b:=462;p:=3;c:=-1;Int((x^(c+2*p)*(1-x)^(2*p)(a+b*x^2))/((a-b)*2^(p-2)*((-‌1)^(c)*(1+x^2))),x=0...1)

22/7 - Pi -> a:=1;b:=0;p:=2;c:=0;Int((x^(c+2*p)*(1-x)^(2*p)*(a+b*x^2))/((a-b)*2^(p-2)*((-1)‌^(c)*(1+x^2))),x=0...1)

Obviously, parameters in the formula somehow depend on "n". The most straight forward dependency on "n" is observed in what Matt B named as "m'" (and I call it "p"): {2,3,4,4,6,6} ... Note that when "n" -> infinity, then the integral should come to 0 ...

Thomas Baruchel has conducted extensive calculations and observed that in originally supplied five parameter notation (i,j,k,l,m)

j = m/2 - 2 and corespondingly m=2*(j+2)

This makes the original conjecture to depend on 4 parameters and to look like:

${\displaystyle (-1)^{n}\cdot (\pi -{\text{A002485}}(n)/{\text{A002486}}(n))=(|i|\cdot 2^{j})^{-1}\int _{0}^{1}{\big (}x^{l}(1-x)^{2(j+2)}(k+(i+k)x^{2}){\big )}/(1+x^{2})\;dx}$

and in unformatted form:

(-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)

This observation confirms previous Matt B result, which also is based on 4 parameters.

Note that expanded expression, presented under the integral on the right hand side consists of 3 terms and looks like:

(k*(1-x)^(2*(j+2))*x^(l+2))/(x^2+1) + (k*(1-x)^(2*(j+2))*x^l)/(x^2+1) + (i*(1-x)^(2*(j+2))*x^(l+2))/(x^2+1)

Update#2:

Based on his calculations results, Thomas Baruchel also found that even with 4 parameters, this formula yields an infinite number of solutions for each n.

And, of course, it would be nice to find some interesting rule on (l,m) in order to minimize (i,k) and, even better, it would be real cool to reduce the number of parameters just to one.

Thomas shared with me his calculations results and so now I have a lot of experimentally found five-tuples {n,i, j, k, l} - where n varies in the range from 2 to 26 and for each value of "n" Thomas supplied me with quite a few of valid combinations of i, j, k, l values ... and based on this data, of course, it would be nice to find how (if at all) i, j, k, l are inter-related between each other and with "n" - but such inter-relation (if exists) is not obvious and difficult to derive just by observation ... (though it is clearly seen that an absolute value of "i" is strongly increasing as "n" is growing from 2 to 26).

Will Sawin noticed (and used in his prove provided in MathOverflow)

that in all found cases [j-l] is congruent modulo 2

If I did not make a mistake RHS could be reduced (after performing integration) to:

(abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HPFQ(1,l/2+1/2,l/2+1;j+l/2+3,  j+l/2+7/2;-1)/Gamma(2*j+l+6))


where HPFQ is the abbreviation for HypergeometricPFQ

May be from discussed parametric identity one could derive irrationality measure for pi, if to assume that RHS in this identity holds true, when the rational fraction on the LHS is equal to 0, then we have:

Pi = (abs(i)*2^j)^(-1)*Gamma(2*j+5)*((k+i)*Gamma(l+3)*HypergeometricPFQ(1,l/2+3/2,l/2+2;j+l/2+4,j+l/2+9/2;-1)/Gamma(2*j+l+8)+k*Gamma(l+1)*HypergeometricPFQ(1,l/2+1/2,l/2+1;j+l/2+3, j+l/2+7/2;-1)/Gamma(2*j+l+6))

Perhaps someone could programmatically check if there are any {i,j,k,l}, which would satisfy above?

Update #3:

Thanks to Jaume Oliver Lafont, at least one case, answering affirmatively to the last question, is identified: i=-1, j=-2, k=1, l=0

${\displaystyle \pi =\int _{0}^{1}{\frac {4}{1+x^{2}}}\,\mathrm {d} x}$

Should there be an infinite number of such cases?

P.S. Per discussion with Jaume Oliver Lafont, depending on the value of the polynomial x degree in the integral body's numerator (while denominator stays to be the same "1+x^2"), the result varies from "Pi" to "log(2)" and also to "+/- (Pi - p/q)" as well as to "+/-(log(2)-p/q)", so perhaps now one could produce two distinct families of parameterization: one for Pi and the differences between Pi and its convergents and another for log(2) and the differences between log(2) and its convergents.

P.P.S. Integrate[x^(2*(j+2))*(1-x)^l*(k+(k+m)*x^2)/((1+x^2)*(Abs[m]*2^j)),{x,0,1}]

yields the following:

ConditionalExpression[(1/Abs[m])2^(-5-3 j-l) Sqrt[\[Pi]] Gamma[5+2 j] Gamma[1+l] (k HypergeometricPFQRegularized[{1,5/2+j,3+j},{3+j+l/2,7/2+j+l/2},-1]+1/2 (3+j) (5+2 j) (k+m) HypergeometricPFQRegularized[{1,7/2+j,4+j},{4+j+l/2,9/2+j+l/2},-1]),Re[l]>-1&&Re[j]>-(5/2)] (2)

Free WolframAlpha timed out and I used as a workaround breaking the integration into two integrals. Integrate[x^(2*(j+2))*(1-x)^l*(k+(k+m)*x^2)/((1+x^2)),{x,0,1}] = Integrate[x^(2*(j+2))*(1-x)^l*k/((1+x^2)),{x,0,1}] + Integrate[x^(2*(j+2))*(1-x)^l*(k+m)*x^2/((1+x^2)),{x,0,1}]

The first (inputted into WolframAlpha in the Maple format)

Int(x^(2*(j+2))*(1-x)^l*(k+m)*x^2/(1+x^2),x=0...1)

yielded

ConditionalExpression[(k Gamma[5 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1])/Gamma[6 + 2 j + l], Re[l] > -1 && Re[j] > -5/2]

The second part

Int(x^(2*(j+2))*(1-x)^l*(k+m)*x^2/(1+x^2),x=0...1)

still was timing out, so I resolved it in Wolfram Cloud Development Platform.(using default Mathematica language)

Integrate[x^(2*(j+2))*(1-x)^l*(k+m)*x^2/(1+x^2),{x,0,1}]

where it gave

ConditionalExpression[((k+m) Gamma[7+2 j] Gamma[1+l] HypergeometricPFQ[{1,7/2+j,4+j},{4+j+l/2,9/2+j+l/2},-1])/Gamma[8+2 j+l],Re[l]>-1&&Re[j]>-(7/2)]

So combining both results we have (without showing conditions)

(k Gamma[5 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1])/Gamma[6 + 2 j + l]+((k+m) Gamma[7+2 j] Gamma[1+l] HypergeometricPFQ[{1,7/2+j,4+j},{4+j+l/2,9/2+j+l/2},-1])/Gamma[8+2 j+l] (3)

I found (perhaps due to conditions) that both 2) and 3) give correct results for the cases: 1), 3) and 5) but for the cases 2) and 4) gave other than Maple tested results (though in both cases the results yielded by both (2) and (3) were the same).

Also in comparing results (2) and (3), it is peculiar to notice that (2) contains explicitly Sqrt[\[Pi]] while (3) doesn't...

I have equated (2) and (3) and resolved it with regards to Sqrt[Pi]

Sqrt[Pi]  = (1/(2^j)*((k Gamma[5 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2,7/2 + j + l/2}, -1])/Gamma[6 + 2 j + l] + ((k + m) Gamma[7 + 2 j] Gamma[1 + l] HypergeometricPFQ[{1, 7/2 + j, 4 + j}, {4 + j + l/2,9/2 + j + l/2}, -1])/Gamma[8 + 2 j + l]))/(2^(-5 - 3 j -l) Gamma[5 + 2 j] Gamma[1 + l] (k HypergeometricPFQRegularized[{1, 5/2 + j,3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1] +1/2 (3 + j) (5 + 2 j) (k + m) HypergeometricPFQRegularized[{1,7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1]))      (4)


which indeed gave Sqrt[Pi] for each set of {j,k,l,m} given in above-listed cases 1), 2), 3), 4), 5)

I presume that above expression (4) will yield Sqrt[Pi] for other (infinite) number of sets of {j,k,l,m}.

Could someone explain my results?

Is it an interesting identity?

I was notified that Maple simplifies the expression on the right hand side of (4) to sqrt(Pi).

It seems to be true for arbitrary j,k,l,m.

Based on stated above that this identity is true for any arbitrary set of {j,k,l,m} we could consider the case j=k=l=m:

Sqrt[Pi] = (2^(5+3 j) (Gamma[5+2 j] Gamma[8+3 j] HypergeometricPFQ[{1,5/2+j,3+j},{3+(3 j)/2,7/2+(3 j)/2},-1]+2 Gamma[7+2 j] Gamma[6+3 j] HypergeometricPFQ[{1,7/2+j,4+j},{4+(3 j)/2,9/2+(3 j)/2},-1]))/(Gamma[5+2 j] Gamma[6+3 j] Gamma[8+3 j] (HypergeometricPFQRegularized[{1,5/2+j,3+j},{3+(3 j)/2,7/2+(3 j)/2},-1]+(15+11 j+2 j^2) HypergeometricPFQRegularized[{1,7/2+j,4+j},{4+(3 j)/2,(3 (3+j))/2},-1])) (5)

References:

###### =====================================================
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 as:

(*) m = n!*omega(n) ] for {n,0,3}


where the function "omega(n)" gives the number of distinct primes in n (in Wolfram notations this function is PrimeNu(n) ).

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

${\displaystyle a(n)=eulerphi(prime(mod(4,n)!!))+floor(n/5)+((1+floor(n/5))^{2})!*((1+sqrt(3))^{(}n-(1+2*(floor(n/5)))!)-(1-sqrt(3))^{(}n-(1+2*(floor(n/5)))!))/(2*sqrt(3)}$

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}

###### ==========================================
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))

"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)))"


###### ==========================================

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

###### ========================================

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

###### ========================================
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)

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

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BBP formula for Pi in a slight disguise

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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)

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From my correspondence with Tito Piezas

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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