A277060 a(n) = (1/2) * Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k+1))^2 for n >= 0.

0, 1, 28, 729, 19376, 529575, 14835780, 424231465, 12338211520, 363931754949, 10862528888300, 327501958094003, 9959845931792784, 305175084350065267, 9412306255856822388, 291982561878565118025, 9104382992541189221120

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

Cf. 1/2 * Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k+1))^m: A050151 (m=1), this sequence (m=2).

Cf. A005259, A074635.

Sequence in context: A004293 A012808 A226991 * A229463 A097834 A162830

Adjacent sequences: A277057 A277058 A277059 * A277061 A277062 A277063

Keyword

nonn,easy

Author

Seiichi Manyama, Nov 07 2016

Status

approved