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)
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))^(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))/(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
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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[km,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(n1))*(floor(sqrt(n1))+1)),n=1..4)),k=0..infinity)
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