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 A179089 a(n) = (Sum_{k=0..n-1} (2k+1)T_k^2(-3)^(n-1-k))/n^2, where T_0, T_1, ... are central trinomial coefficients given by A002426. 2
 1, 0, 5, 13, 105, 576, 4005, 27000, 193193, 1402672, 10433709, 78807785, 603996745, 4683970032, 36702939429, 290184446349, 2312460578025, 18556825469040, 149842592021997, 1216719520281045, 9929612901775761, 81406058258856240 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS 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,... LINKS G. C. Greubel, Table of n, a(n) for n = 1..1000 Zhi-Wei Sun, Arithmetic properties of Apery numbers and central Delannoy numbers, preprint, arXiv:1006.2776 [math.NT], 2010-2011. EXAMPLE For n=4 we have a(4) = (T_0^2(-3)^3 + 3T_1^2(-3)^2 + 5T_2^2(-3) + 7T_3^2)/4^2 = (-27 + 27 - 5*27 + 7^3)/16 = 13. MATHEMATICA 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}] CROSSREFS Cf. A002426, A178808, A178790, A178791, A173774. Sequence in context: A305912 A180827 A117437 * A106046 A180362 A117527 Adjacent sequences:  A179086 A179087 A179088 * A179090 A179091 A179092 KEYWORD nonn AUTHOR Zhi-Wei Sun, Jun 29 2010 STATUS approved

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Last modified September 20 05:58 EDT 2020. Contains 337264 sequences. (Running on oeis4.)