Alex
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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