OFFSET
0,1
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) ).
FORMULA
EXAMPLE
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).
MAPLE
seq(add( binomial(n+k-1, k) + binomial(n+k-1, k)^2, k = 0..n ), n = 0..20);
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Oct 10 2022
STATUS
approved