A357671 a(n) = Sum_{k = 0..n} ( binomial(n+k-1,k) + binomial(n+k-1,k)^2 ).

2, 4, 20, 166, 1812, 22504, 297362, 4067298, 56897300, 809019580, 11649254520, 169444978124, 2485270719570, 36707044807996, 545386321069862, 8144809732228666, 122177690210103060, 1839933274439787940, 27804610626798500372, 421476329345312885304, 6406685025104178888312

Offset: 0

Author

Peter Bala, Oct 10 2022

Keywords

• nonn
• easy

Comments

Conjectures:

1) a(p) == 4 (mod p^5) for all primes p >= 7 (checked up to p = 499). Note that A000984(p) == 2 (mod p^3) and A333592(p) == 2 (mod p^3) for all primes p >= 5.

2) For r >= 2, and all primes p >= 5, 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) = 2*m*Sum_{k = 0..m*n} binomial(n+k-1,k) + (m + 1)*Sum_{k = 0..m*n} binomial(n+k-1,k)^2. Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 1. See A357673 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(19) - a(1) = 421476329345312885304 - 4 = (2^2)*(5^2)*(19^5)*1913*2383*373393 == 0 (mod 19^5).
a(25) - a(5) = 5375188503768910125546897504 - 22504 = (2^3)*(5^10)*1858537* 37019662696111 == 0 (mod 5^10).
		

Crossrefs

Cf. A000984, A333592, A357509, A357565, A357566, A357672, A357673, A357674.

Programs

Maple

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

PARI

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

Python

from math import comb
def A357671(n): return comb(n<<1,n)+sum(comb(n+k-1,k)**2 for k in range(n+1)) if n else 2 # Chai Wah Wu, Oct 28 2022

Formula

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

a(p) == 4 (mod p^3) for all primes p >= 5.

a(n) ~ 2^(4*n) / (3*Pi*n).