I made a conjecture that:
(1)^n*(Pi−A002485(n)/A002486(n))=(abs(i)2^j)^(1)Int((x^l(1x)^(2(j+2))*(k+(i+k)*x^2))/(1+x^2),x=0...1)
A002485(n) & A002486(n) are "n" indexed OEIS.org integer sequences and {"i", "j", "k", "l"} are sets of integers to be found for the each value of "n". Could above be proved analytically showing that for each "n" an infinite number of above sets could be found. Is there a relationship between some 2 parameters in those sets, so 1 of them could be eliminated reducing the number of those from 4 parameters to 3?
I found the identity 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]))
It's true for any arbitrary sets of {j,k,l,m} signed integers as confirmed by Mathematica's WolframAlpha & Maple.
=================================== I found the identity
sqrt(exp(1)) = 16/31*(sum((1/2)^n*(1/2n^3+1/2n+1)/n!,n=1..infinity) +1)
or
sqrt(e) = (16/31)(1 + Sum_{n>=1}(1/2)^n(1/2n^3+1/2n+1)/n!)
http://www.strw.leidenuniv.nl/~mathar/public/mathar20071105.pdf https://oeis.org/A019774
I suggested two formulas for Heegner numbers (see OEIS A003173):
a) for the first four (smallest) Heegner numbers
a(n) = 1+((1 + sqrt(3))^(n1)  (1  sqrt(3))^(n1))/(2*sqrt(3)) for n = 1,2,3,4
b) for the last (largest) four Heegner numbers
a(n) = 19+24*((1 + sqrt(3))^(n6)  (1  sqrt(3))^(n6))/(2*sqrt(3)) for n = 6,7,8,9
In general
a(n) = a(k) + (a(k+1)a(k))((1 + sqrt(3))^(nk)  (1  sqrt(3))^(nk))/(2sqrt(3)) where for n =1,2,3,4 k=1 and for n =6,7,8,9 k=6
Three hard to prove conjectures from me (Alexander R. Povolotsky)
n! + prime(n) != m^k (so far proven only for the case when k=2)
n! + n^2 != m^2 (so far proven only for the case when n is prime number)
n! + Sum(j^2, j=1, j=n) != m^2 (so far no proof) where != means "not equal" and k,m,n are integers
7901234568 / 9876543210 * 1234567890 = 0987654312
24/Pi = sum((30k+7)binom(2k,k)^2(Hypergeometric2F1[1/2  k/2, k/2, 1, 64])/(256)^k, k=0...infinity) or Sum[(30k+7)Binomial[2k,k]^2(Sum[Binomial[km,m]*Binomial[k,m]*16^m,{m,0,k/2}])/(256)^k,{k,0,infinity}]
BBP formula in a disguise sum((1/16)^k*(sum(((1)^(ceil(4/(2n))))(floor(4/n))/(8k+n+floor(sqrt(n1))(floor(sqrt(n1))+1)),n=1..4)),k=0..infinity)
Pi^2 = lim (n*(n+1)(2n+1))*((sum(1/i^2,i=1...n))/(sum(i^2,i=1...n))), n>infinity

YearlingNov 14, 2019

CommentatorJul 14, 2014

EditorJul 12, 2014

TeacherAug 1, 2014