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A178791
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The arithmetic mean of (2k+1)(-1)^k*A_k (k=0,...,n-1), where A_0, A_1,... are Apéry numbers given by A005259.
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7
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1, -7, 117, -2441, 57449, -1453635, 38609845, -1061792695, 29973352185, -863536596143, 25288254409373, -750531594051981, 22525211241191881, -682459907754004723, 20845409947239778533, -641211780685502724425
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OFFSET
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1,2
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COMMENTS
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On Jun 14 2010, Zhi-Wei Sun conjectured that the number a(n) = n^{-1}*Sum_{k=0..n-1} (2*k+1)*(-1)^k*A_k is always an integer and that a(p) = p(p/3) (mod p^3) for any prime p>3. He also formulated the following conjecture: If p=1 (mod 3) is a prime and p = x^2 + 3y^2 with x,y integers then Sum_{k=0..p-1} (-1)^k*A_k = 4*x^2 - 2*p (mod p^2); if p=2 (mod 3) is a prime then Sum_{k=0..p-1 }(-1)^k*A_k=0 (mod p^2).
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LINKS
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FORMULA
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G.f.: apart from the minus signs (just replace x by -x) the generating function is 2*x*G/(x+1) + Int((x-1)*G/(x+1)^2, x) where G is the generating function of A005259. - Mark van Hoeij, May 07 2013
a(n) ~ -(-1)^n * 2^(3/4) * (1 + sqrt(2))^(4*n) / (24 * (Pi*n)^(3/2)). - Vaclav Kotesovec, Jan 24 2019
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EXAMPLE
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For n=3 we have a(3) = (A_0 - 3A_1 + 5A_2)/3 = (1 - 3*5 + 5*73)/3 = 117.
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MAPLE
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G := (-1/2)*(3*x-3+(x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3, 2/3], [1], (-1/2)*(x^2-7*x+1)*(x+1)^(-3)*(x^2-34*x+1)^(1/2)+(1/2)*(x^3+30*x^2 -24*x+1)*(x+1)^(-3))^2;
ogf := 2*x*G/(x+1)+Int((x-1)*G/(x+1)^2, x);
series(ogf, x=0, 25);
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MATHEMATICA
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Apery[n_]:= Sum[Binomial[n+k, k]^2Binomial[n, k]^2, {k, 0, n}]; AA[n_]:= Sum[(2k+1)(-1)^k*Apery[k], {k, 0, n-1}]/n; Table[AA[n], {n, 25}]
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PROG
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(PARI) A(n) = sum(k=0, n, (binomial(n, k)*binomial(n+k, k))^2); \\ A005259
a(n) = sum(k=0, n-1, (2*k+1)*(-1)^k*A(k))/n; \\ Michel Marcus, Jan 24 2019
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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