A357565 a(n) = 3*Sum_{k = 0..n} binomial(n+k-1,k)^2 + 2*Sum_{k = 0..n} binomial(n+k-1,k)^3.

5, 10, 114, 2926, 109106, 4846260, 234488526, 11913003294, 625130924082, 33590792825200, 1838547540484364, 102135528447552060, 5743779960435245774, 326352202770939600460, 18706076476872783254286, 1080345839256279791104926, 62806507721442655949609010

Offset: 0

Author

Peter Bala, Oct 16 2022

Keywords

• nonn
• easy

Comments

Conjectures:

1) a(p) == a(1) (mod p^5) for all odd primes p except p = 5 (checked up to p = 271).

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

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

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

Examples

			a(11) - a(1) = 102135528447552060 - 10 = 2*(5^2)*(11^5)*14657* 865363 == 0 (mod 11^5).
a(5^2) - a(5) = 581553752659150682384860284864053981408760 - 4846260 = 3*(2^2)*(5^9)*5611847956825027*4421531072180960789 == 0 (mod 5^9)
		

Crossrefs

Cf. A357566, A357671, A357672, A357673, A357674.

Programs

Maple

seq(add( 3*binomial(n+k-1,k)^2 + 2*binomial(n+k-1,k)^3, k = 0..n ), n = 0..20);

PARI

a(n) = 3*sum(k = 0, n, binomial(n+k-1,k)^2) + 2*sum(k = 0, n, binomial(n+k-1,k)^3); \\ Michel Marcus, Oct 25 2022