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

where A002485(n) and A002486(n) are integer sequences indexed by "n" and referenced in the OEIS.org and "i", "j", "k", "l" - are some integers, together representing the set, and to be experimentally found separately for the each value of "n"... Could my conjecture be proved analytically? PS There is a alleged possibility, that for each "n" an infinite number of {"i", "j", "k", "l"} sets could be found... Is there a relationship between some two parameters in a {"i", "j", "k", "l"} sets, so one of parameters could be eliminated thus reducing the number of parameters from 4 to 3?

I found the following 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 seems to be true for any arbitrary sets of {j,k,l,m} signed integers as confirmed by both Mathematica based WolframAlpha and Maple.

I suggested the following two formulas for Heegner numbers (see OEIS A003173):

a) for the first four (smallest) Heegner numbers

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

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

In general

## a(n) = a(k) + (a(k+1)-a(k))((1 + sqrt(3))^(n-k) - (1 - sqrt(3))^(n-k))/(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 Alexander R. Povolotsky

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

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

## 7901234568 / 9876543210 * 1234567890 = 0987654312

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

or

## BBP formula in a slight disguise sum((1/16)^k*(sum(((-1)^(ceil(4/(2n))))(floor(4/n))/(8k+n+floor(sqrt(n-1))(floor(sqrt(n-1))+1)),n=1..4)),k=0..infinity)

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

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