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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))^(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}]

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