A278415 -id:A278415 - cp's OEIS frontend

A278405 a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).

1, 1, 2, 19, 110, 476, 2477, 15093, 86830, 485290, 2826902, 16857116, 100034453, 594833357, 3574477090, 21611465819, 130955824174, 796195223398, 4860425688176, 29760574848750, 182655048136510, 1123720751229858, 6929124085148938, 42811398244528788

Offset: 0

Zhi-Wei Sun, Nov 20 2016

nonn

Conjecture: For any prime p > 5 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.
We have proved that for any prime p > 5 and positive integer n the number (a(p*n)-a(n))/(p^3*n^2) is always a p-adic integer.
Diagonal of the rational function 1 / ((1 + x)*(1 - x)*(1 - y)*(1 - z) - x*y*z). - Ilya Gutkovskiy, Apr 23 2025

			a(3) = 19 since a(3) = C(3,2*0)^2*C(3-0,0) + C(3,2*1)^2*C(3-1,1) = 1 + 3^2*2 = 19.
G.f. = 1 + x + 2*x^2 + 19*x^3 + 110*x^4 + 476*x^5 + 2477*x^6 + 15093*x^7 + ...

Zhi-Wei Sun, Table of n, a(n) for n = 0..200
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Cf. A208425, A244973, A275027, A277640, A278415.

a[n_]:=a[n]=Sum[Binomial[n,2k]^2*Binomial[n-k,k],{k,0,n/2}]
Table[a[n],{n,0,27}]

cp's OEIS Frontend

A278405 a(n) = Sum_{k=0..n} binomial(n,2k)^2*binomial(n-k,k).

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