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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 (but not proved) possibility, that for each "n" an infinite number of {"i", "j", "k", "l"} sets could be found... - so the question arises whether there is a relationship between some two parameters in a {"i", "j", "k", "l"} sets, so one of those parameters could then be eliminated from the conjectured by me formula, thus reducing the number of parameters (members in the set) in it from 4 to 3?

I suggest 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))/(2*sqrt(3)) where for n =1,2,3,4 k=1 and for n =6,7,8,9 k=6

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

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)

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}]

BBP formula in a slight disguise 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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