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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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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 more information on this at

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

The number of "PovolotskyPairs", calculated using the program, written by R. J. Cano,

(for bases from to 2 to 10 - see answer section of the above StackExchange link) is:

{2,2,3,3,5,3,7,5,7,...} for n>=1 where n=r-1 and r is the base radix.

Judging by above sequence it appears that the number of "PovolotskyPairs" is seems to be related to phi, which is the Euler totient function - see A039649, A039650, A214288.

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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

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

(-1)^n*(π−A002485(n)/A002486(n))=(|i|⋅2j)−1∫10(xl(1−x)m(k+(i+k)x2))/(1+x2)dx

where integer n = 0,1,2,3,... serves as 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)

n=3, i=-1, j=0, k=1, l=4, m=4

In Lucas's formula for 333/106 (mentioned above in the comment by Chandrasekhar)

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

n=4, i=265, j=1, k=197, l=5, m=6

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)

n=5, i=791, j=2, k=25, l=8, m=8

In Lucas's formula for 103993/33102

Pi - 103993/33102 = 1/75521*Int(x^14*(1-x)^12*(124360-77159*x^2)/(1+x^2),x = 0 .. 1)

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

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)

n=7, i= -2409, j=4, k=1349, l=12, m=12

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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 itselfvs its 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), 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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