A357672 a(n) = Sum_{k = 0..n} binomial(n+k-1,k) * Sum_{k = 0..n} binomial(n+k-1,k)^2.

1, 4, 84, 2920, 121940, 5607504, 273908712, 13947188112, 732102614100, 39332168075200, 2152235533317584, 119531412173662944, 6720552415489860584, 381775182057562837600, 21879043278489630349200, 1263402662473729731877920, 73438613319490294002441300, 4293679728171938162242298400

Offset: 0

Author

Peter Bala, Oct 10 2022

Keywords

• nonn
• easy

Comments

Conjectures:

1) a(p) == 4 (mod p^5) for all odd primes p except p = 5 (checked up to p = 499). Note that both A000984(p) == 2 (mod p^3) and A333592(p) == 2 (mod p^3) for all primes p >= 5 and hence a(p) == 4 (mod p^3) for all primes p >= 5.

2) For r >= 2, and all primes p >= 3, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).

3) More generally, let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k) )^(2*m) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+1). Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. See A357674 for the case m = 2.

4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).

Examples

			Examples of supercongruences:
a(17) - a(1) = 4293679728171938162242298400 - 4 = (2^2)*(17^5)*3457* 218688360593678551 == 0 (mod 17^5).
a(5^2) - a(5) = (2^4)*(3^2)*(5^9)*7*7229*102559*465516030080883405648119 == 0 (mod 5^9).
		

Crossrefs

Cf. A000984, A333592, A357565, A357566, A357671, A357673, A357674.

Programs

Maple

seq(add(binomial(n+k-1,k), k = 0..n) * add( binomial(n+k-1,k)^2, k = 0..n), n = 0..20);

PARI

a(n) = sum(k = 0, n, binomial(n+k-1,k)) * sum(k = 0, n, binomial(n+k-1,k)^2); \\ Michel Marcus, Oct 24 2022

Formula

a(n) = A000984(n) * A333592(n).

a(n) ~ 2^(6*n)/(3*(Pi*n)^(3/2)).