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A179089
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a(n) = (1/n^2) * Sum_{k=0..n-1} (2k+1)*T_k^2(-3)^(n-1-k), where T_0, T_1, ... are central trinomial coefficients given by A002426.
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2
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1, 0, 5, 13, 105, 576, 4005, 27000, 193193, 1402672, 10433709, 78807785, 603996745, 4683970032, 36702939429, 290184446349, 2312460578025, 18556825469040, 149842592021997, 1216719520281045, 9929612901775761, 81406058258856240
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OFFSET
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1,3
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COMMENTS
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On Jun 28 2010, Zhi-Wei Sun conjectured that a(n) is an integer for every n=1,2,3,... and that a(p) = (1+p/3)/2 (mod p) for any prime p, where (p/3) is the Legendre symbol. In contrast, he showed that Sum_{k=0..n-1} (2k+1)*T_k^2*3^(n-1-k) = n*T_n*T_{n-1} for all n=1,2,3,...
The formula for a(n) in the formula section implies that a(n) is an integer. - Mark van Hoeij, Nov 13 2022
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LINKS
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FORMULA
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G.f.: Integral(hypergeom([1/2, 3/2], [2], 16*x/(1 + 3*x)^2)/(1 + 3*x)^2). - Mark van Hoeij, Nov 10 2022
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EXAMPLE
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For n = 4 we have a(4) = (T_0^2(-3)^3 + 3*T_1^2(-3)^2 + 5*T_2^2(-3) + 7*T_3^2)/4^2 = (-27 + 27 - 5*27 + 7^3)/16 = 13.
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MAPLE
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MATHEMATICA
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TT[n_]:=Sum[Binomial[n, 2k]Binomial[2k, k], {k, 0, Floor[n/2]}] SS[n_]:=Sum[(2k+1)*TT[k]^2*(-3)^(n-1-k), {k, 0, n-1}]/n^2 Table[SS[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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