A179089 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.

1, 0, 5, 13, 105, 576, 4005, 27000, 193193, 1402672, 10433709, 78807785, 603996745, 4683970032, 36702939429, 290184446349, 2312460578025, 18556825469040, 149842592021997, 1216719520281045, 9929612901775761, 81406058258856240

Offset: 1

Author

Zhi-Wei Sun, Jun 29 2010

Keywords

• nonn

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,...

The formula for a(n) in the formula section implies that a(n) is an integer. - Mark van Hoeij, Nov 13 2022

Examples

			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.
		

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.

Crossrefs

Cf. A002426, A178808, A178790, A178791, A173774.

Programs

Maple

A002426 := n -> simplify(GegenbauerC(n, -n, -1/2)); seq( (A002426(n)+A002426(n-1))*(3*A002426(n-1)-A002426(n))/4, n=1..20); # Mark van Hoeij, Nov 13 2022

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}]

Formula

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

a(n) = (A002426(n)+A002426(n-1))*(3*A002426(n-1)-A002426(n))/4. - Mark van Hoeij, Nov 13 2022