A329475 a(n) = Sum_{k=0..n} C(n,k)^2*T(k)*T(n-k), where T(k) = A002426(k) is the coefficient of x^k in the expansion of (x^2+x+1)^k.

1, 2, 10, 68, 586, 5252, 49204, 475400, 4723786, 47937812, 494786260, 5177188040, 54794164660, 585565913480, 6309889976680, 68484312535568, 747985368753226, 8214968193003860, 90669516557975524, 1005156080857529768, 11187435500257898836, 124964856185950621832

Offset: 0

Author

Zhi-Wei Sun, Nov 13 2019

Keywords

• nonn

Comments

The author introduced this sequence in arXiv:1911.05456 and made the following conjecture.

Conjecture: Let p be an odd prime and let S = Sum_{k=0..p-1}a(k)/(-4)^k. If p == 1 (mod 12) and p = x^2 + 9*y^2 with x and y integers, then S == 4*x^2-2*p (mod p^2). If p == 5 (mod 12) and p = x^2 + y^2 with x == y (mod 3), then S == 4*x*y (mod p^2). If p == 3 (mod 4), then S == 0 (mod p^2).

Note that if p > 3 is a prime, then a(p-1) == Sum_{k=0..p-1} T(k)*T(p-1-k) == Legendre(p/3)*Sum_{k=0..p-1}T(k)^2/(-3)^k == 1 (mod p) by (1.7) and (2.3) of the author's 2014 paper in Sci. China Math.

Examples

			a(1) = 2 since Sum_{k=0,1} C(1,k)^2*T(k)*T(1-k) = C(1,0)^2*T(0)*T(1) + C(1,1)^2*T(1)*T(0) = 2*T(0)*T(1) = 2*1*1 = 2.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..931 (terms 0..150 from Zhi-Wei Sun)
• Zhi-Wei Sun, Congruences involving generalized central trinomial coefficients, Sci. China Math. 57(2014), no.7, 1375-1400.
• Zhi-Wei Sun, On sums related to central binomial and trinomial coefficients, in: M. B. Nathanson (ed.), Combinatorial and Additive Number Theory: CANT 2011 and 2012, Springer Proc. in Math. & Stat., Vol. 101, Springer, New York, 2014, pp. 257-312. Also available from arXiv:1101.0600 [math.NT], 2011-2014.
• Zhi-Wei Sun, New series for powers of Pi and related congruences, Electron. Res. Arch. 28 (2020), no. 3, 1273--1342.
• Zhi-Wei Sun, On central trinomial coefficients, Question 491563 at MathOverflow, April 23, 2025.

Crossrefs

Cf. A002426, A002895.

Programs

Mathematica

T[0]=1; T[1]=1; T[n_]:=T[n]=((2n-1)T[n-1]+3*(n-1)*T[n-2])/n;
a[n_]:=a[n]=Sum[Binomial[n,k]^2*T[k]*T[n-k],{k,0,n}];
Table[a[n],{n,0,21}]

Formula

a(n) ~ (3/2)*12^n/(n*Pi)^(3/2) as n tends to the infinity.