OFFSET
0,2
COMMENTS
Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 271).
2) For r >= 2, and all primes p >= 3, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
2) 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. This is the case m = 2. See A357672 for the case m = 1.
FORMULA
a(n) = ( A005809(n) )^4 * (Sum_{k = 0..2*n} binomial(n+k-1,k)^2 )^3.
EXAMPLE
Example of a supercongruence:
a(7) - a(1) = 4181198339699286332943143923058721957212160000000 - 2187 = (3^7)*(7^5)*211*298225180113209*1807736060307048120859243 == 0 (mod 7^5).
MAPLE
seq((add(binomial(n+k-1, k), k = 0..2*n))^4 * (add( binomial(n+k-1, k)^2, k = 0..2*n))^3, n = 0..20);
PROG
(PARI) a(n) = sum(k = 0, 2*n, binomial(n+k-1, k))^4 * sum(k = 0, 2*n, binomial(n+k-1, k)^2)^3; \\ Michel Marcus, Oct 24 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Oct 11 2022
STATUS
approved