OFFSET
0,3
COMMENTS
Conjecture: the supercongruences a(p-1) == 1 (mod p^4) holds for all primes p >= 5 and a(p^2-1) == 1 (mod p^5) holds for all primes p >= 3. - Peter Bala, Mar 22 2023
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..656
FORMULA
a(n) = n^2 * A074635(n)/2.
From Peter Bala, Mar 22 2023: (Start)
a(n) = Sum_{k = 0..n-1} binomial(n+1,k)*binomial(n-1,k)*binomial(n+k,k)^2.
P-recursive: (n-1)^2*(3*n^2-6*n+2)*(n+1)^3*a(n) = (2*n-1)*(51*n^4-102*n^3+19*n^2+ 32*n-14)*n^2*a(n-1) - n^2*(n-2)*(3*n^2-1)*(n-1)^2*a(n-2) with a(0) = 0 and a(1) = 1.
a(n) ~ sqrt(12 + 17*sqrt(2)/2)*(17 + 12*sqrt(2))^n/(4*n^(3/2)*Pi^(3/2)). (End)
MAPLE
a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 else ( (2*n-1)*(51*n^4-102*n^3+19*n^2+ 32*n-14)*n^2*a(n-1) - n^2*(n-2)*(3*n^2-1)*(n-1)^2*a(n-2) )/( (n-1)^2*(3*n^2-6*n+2)*(n+1)^3 ) end if; end:
seq(a(n), n = 0..20); # Peter Bala, Mar 22 2023
PROG
(PARI) a(n)=my(t=n); if(n<2, return(n)); sum(k=1, n, t*=(n-k+1)*(n+k)/k/(k+1); t^2, n^2)/2 \\ Charles R Greathouse IV, Nov 07 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Seiichi Manyama, Nov 07 2016
STATUS
approved