A383274 a(n) = Sum_{i,j = 0..n} C(n, i)^2*C(n, j)^2*C(i+j, i)*2^(i+j).

1, 13, 441, 20629, 1119361, 66116013, 4126228569, 267666251733, 17868312820737, 1219477111897933, 84701899713767161, 5967906378862013973, 425503428034568158081, 30642774518964618986989, 2225692868157573335052441, 162858794856607965831417429, 11993850186156155815298686977

Offset: 0

Author

Zhi-Wei Sun, Apr 26 2025

Keywords

• nonn

Comments

Conjecture 1: Let S(p) = Sum_{0

If p == 1,9 (mod 20) with p = x^2 + 5*y^2, then S(p) == 4*x^2-2*p (mod p^2).

If p == 3,7 (mod 20) with 2*p = x^2 + 5*y^2, then S(p) == 2*x^2 - 2*p (mod p^2).

If p == 11, 13, 17, 19 (mod 20), then S(p) == 0 (mod p^2).

Conjecture 2: For any prime p == 1,3,7,9 (mod 20), we have Sum_{0

Both conjectures have been verified for odd primes smaller than 1000.

Examples

			a(1) = 13 since Sum_{i,j = 0,1}C(1,i)^2*C(1,j)^2*C(i+j,i)^2*2^(i+j) = Sum_{i,j = 0,1} C(i+j,i)^2*2^(i+j) = 2^0 + 2^1 + 2^1 + C(2,1)*2^2 = 13.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..85
• Zhi-Wei Sun, On a_n(x) = Sum_{i,j=0..n}C(n,i)^2*C(n,j)^2*C(i+j,i)*x^(i+j) (I), Question 491655 at MathOverflow, April 24, 2025.
• Zhi-Wei Sun, A family of polynomials and related congruences and series, arXiv:2505.02767 [math.NT], 2025.

Crossrefs

Cf. A005259.

Programs

Maple

a := proc(n) option remember; local i, j; add(add(binomial(n, i)^2 * binomial(n, j)^2 * binomial(i+j, i) * 2^(i+j), i = 0..n), j = 0..n) end: seq(a(n), n=0..16);  # Peter Luschny, Apr 27 2025

Mathematica

a[n_] := a[n] = Sum[Binomial[n,i]^2*Binomial[n,j]^2*Binomial[i+j,i]*2^(i+j), {i,0,n}, {j,0,n}]; Table[a[n], {n,0,17}]

Python

from math import comb
def A383274(n): return sum((comb(n,i)**2<Chai Wah Wu, Apr 27 2025

Formula

a(n) ~ 3^(4*n+3) / (8 * sqrt(7) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Apr 27 2025

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