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Alexander R. Povolotsky apovolot@gmail.com

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

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 k,m,n are integers

Contents

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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 9876543210 / 1234567890 = 8.0000000729...

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

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)(n2(n + 1)n − (n + 1)n + 1) / ( − n2 + n(n + 1)n + (n + 1)nn − 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

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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

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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:

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 ?

PS See also

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

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

I conjecture that the following identity below represents a generalization of Stephen Lucas' experimentally obtained identities between Pi and its convergents:

(-1)^n\cdot(\pi - \text{A002485}(n)/\text{A002486}(n))
=(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^m(k+(i+k)x^2)\big)/(1+x^2)\; dx

In unformatted form:

(-1)^n*(π−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 = 0,1,2,3,... 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:

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


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)

\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


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)

\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


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)

\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


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)

\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


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)

\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


And it works as well for \frac{618669248999119}{196928538206400}

 \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 $n=6, i= 47201, j=4, k=77159, l=14, m=12$ -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


See also http://math.stackexchange.com/questions/1956/is-there-an-integral-that-proves-pi-333-106/127618#127618

UPDATE: Per http://math.stackexchange.com/questions/860499/seeking-proof-for-the-formula-relating-pi-with-its-convergents

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

(-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 ...

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On Ramanujan constant and alike
=====================================================

Ramanujan constant and alike could be approximated in general as:

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

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

1) n=0, k= 4 ;

2) n=1, k= 40 ;

3) n=2, k= 220 ;

4) n=6, k = 26680 ; -this of course is the case for Ramanujan constant itself with regards to it integer counterpart approximation

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

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

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

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

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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))

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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))

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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) )

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

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

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

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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

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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)

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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

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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))

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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)

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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.

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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

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

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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

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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)

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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

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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

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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

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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)))

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Infinite sums for Euler number (Napier’s constant) & its roots
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exp(1)= (1+ sum((1+n^(3)+n)/(1^n*n!),n=1…infinity))/7

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

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

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

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

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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

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Approximation of Pi involving e 
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Pi ~= sqrt(4e-1)

which is good to 2 decimal digits

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Approximate identities based on linear combinations of symbolic constants
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Pi~=1/17*(1+50*sqrt(log(3)))

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

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

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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)

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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

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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

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

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(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

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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

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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

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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

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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

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

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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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(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

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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).

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  BBP Log(3) formula

According to "A Compendium of BBP-Type Formulas for Mathematical Constants" by David H. Bailey.

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

see page 24 there where it talks about Formula (62) [25]: "Alexander Povolotsky discovered the formula ..."

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 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

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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

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