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A181147
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a(n) = Sum_{k=0..n-1} (105k+44)*C(2k,k)^2*T(k)*(-1)^(n-1-k)/(2n*C(2n,n)), where T(k) (k=0,1,2,...) are central trinomial coefficients given by A002426.
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1
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11, 23, 224, 1747, 16754, 162392, 1651206, 17126327, 181182446, 1943132842, 21080299228, 230802972664, 2546569337336, 28280754214358, 315824396838386, 3544003431783795, 39936833763112790, 451718158386620678
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OFFSET
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1,1
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COMMENTS
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On Jan 22 2011, Zhi-Wei Sun conjectured that a(n) is a positive integer for every n=1,2,3,... Let p > 3 be a prime. He also conjectured that a(p) == 5 + 6*(p/3)*(2-3^(p-1)) (mod p^2). Another conjecture of his states that Sum_{k=0..p-1} (-1)^k*binomial(2k,k)^2*T(k) is congruent to b(p) modulo p^2, where b(p)=0 if (p/15)=-1, b(p) = 4x^2-2p if p == 1,4 (mod 15) and p = x^2+15y^2 with x,y integers, and b(p) = 20x^2-2p if p == 2,8 (mod 15) and p=5x^2+3y^2 with x,y integers.
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LINKS
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EXAMPLE
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For n=2 we have a(2) = (44*1^2*T(0)(-1) + (105+44)*2^2*T(1))/(2*2*binomial(4,2)) = 23.
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MAPLE
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T:= n-> `if`(n=0, 1, coeff ((x^2+x+1)^n, x, n)):
a:= n-> add ((105*k+44) *binomial (2*k, k)^2 *T(k)*(-1)^(n-1-k),
k=0..n-1)/ (2*n*binomial (2*n, n)):
seq (a(n), n=1..30);
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MATHEMATICA
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T[k_]:=If[k>0, Coefficient[(x^2+x+1)^k, x^k], 1]
A[n_]:=Sum[(105k+44)Binomial[2k, k]^2*T[k](-1)^(n-1-k), {k, 0, n-1}]/(2n*Binomial[2n, n])
Table[A[n], {n, 1, 50}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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