A357673 a(n) = 4*Sum_{k = 0..2*n} binomial(n+k-1,k) + 3*Sum_{k = 0..2*n} binomial(n+k-1,k)^2.

7, 21, 225, 5124, 162657, 5812521, 219004812, 8516056500, 338508840801, 13679415485805, 559978704877725, 23162632151271480, 966309241173439500, 40602415885424806824, 1716435895297948558812, 72941388509291664563124, 3113826813351114598588257, 133458673478315967012049245

Offset: 0

Author

Peter Bala, Oct 11 2022

Keywords

• nonn
• easy

Comments

Conjectures:

1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 499).

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) = 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 = 2. See A357673 for the case m = 1.

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) = 133458673478315967012049245 - 21 = (2^3)*3*7*(17^5)*61*109*4441*86491*219071 == 0 (mod 17^5).
a(25) - a(5) = 1681058690656849873108154414589433546896 - 5812521 = 3*(5^9)*17*124471*39410141*65963867*52155532801 == 0 (mod 5^9).
		

Crossrefs

Cf. A005809, A357509, A357565, A357566, A357671, A357672, A357674.

Programs

Maple

seq(add( 4*binomial(n+k-1,k) + 3*binomial(n+k-1,k)^2, k = 0..2*n ), n = 0..20);

Mathematica

Table[4 Sum[Binomial[n+k-1,k],{k,0,2n}]+3*Sum[Binomial[n+k-1,k]^2,{k,0,2n}],{n,0,20}] (* Harvey P. Dale, Oct 29 2022 *)

PARI

a(n) = 4*sum(k = 0, 2*n, binomial(n+k-1,k)) + 3*sum(k = 0, 2*n, binomial(n+k-1,k)^2); \\ Michel Marcus, Oct 24 2022

Formula

a(n) = 4*A005809(n) + 3*Sum_{k = 0..2*n} binomial(n+k-1,k)^2.