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A179100 a(n) = (1/n) * Sum_{k=0..n-1} (8k+5) T_k^2, where T_0, T_1, ... are central trinomial coefficients given by A002426. 0

%I #11 Aug 04 2019 18:59:31

%S 5,9,69,407,2997,22005,169389,1325889,10573677,85386881,697013325,

%T 5739021051,47599593941,397234035333,3332690347437,28089543969855,

%U 237711099004461,2018856328439841,17200553934626253,146966002696538271

%N a(n) = (1/n) * Sum_{k=0..n-1} (8k+5) T_k^2, where T_0, T_1, ... are central trinomial coefficients given by A002426.

%C On Jun 17 2010, _Zhi-Wei Sun_ conjectured that a(n) is an integer for every n=1,2,3,... and that a(p) == 3(p/3) (mod p) for any prime p, where (p/3) is the Legendre symbol. He also observed that Sum_{k=0..n-1} (2k+1) T_k*3^{n-1-k} = n * Sum_{k=0..n-1} C(n-1,k)*(-1)^(n-1-k)*(k+1)*C(2k,k).

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1006.2776">Arithmetic properties of Apery numbers and central Delannoy numbers</a>, preprint, arXiv:1006.2776 [math.NT], 2010-2011.

%e For n=3 we have a(3) = (5*T_0^2 + 13*T_1^2 + 21*T_2^2)/3 = (5 + 13 + 21*9)/3 = 69.

%t TT[n_]:=Sum[Binomial[n,2k]Binomial[2k,k],{k,0,Floor[n/2]}] SS[n_]:=Sum[(8k+5)*TT[k]^2,{k,0,n-1}]/n Table[SS[n],{n,1,50}]

%Y Cf. A002426, A179089, A178808, A178790, A178791, A173774.

%K nonn

%O 1,1

%A _Zhi-Wei Sun_, Jun 29 2010

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Last modified August 30 19:26 EDT 2024. Contains 375545 sequences. (Running on oeis4.)