OFFSET
0,2
COMMENTS
Conjecture 1: Let S(p) = Sum_{0<k<p} a(k) with p a prime.
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<k<p}(8k+5)*a(k) == 14/5*p (mod p^2).
Both conjectures have been verified for odd primes smaller than 1000.
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.
FORMULA
a(n) ~ 3^(4*n+3) / (8 * sqrt(7) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Apr 27 2025
EXAMPLE
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.
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}]
PROG
(Python)
from math import comb
def A383274(n): return sum((comb(n, i)**2<<i)*sum(comb(n, j)**2*comb(i+j, i)<<j for j in range(n+1)) for i in range(n+1)) # Chai Wah Wu, Apr 27 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 26 2025
STATUS
approved