A176335 Central coefficients T(2n,n) of number triangle A176331.

1, 3, 28, 315, 3876, 50358, 678112, 9365499, 131809060, 1882294128, 27193657008, 396600597198, 5829739893264, 86262567856650, 1283677784658528, 19196304797150715, 288295493121264420, 4346056823245242420

Offset: 0

Author

Paul Barry, Apr 15 2010

Keywords

• nonn
• easy

Links

• G. C. Greubel, Table of n, a(n) for n = 0..825
• F. Baldassarri, S. Bosch, B. Dwork, (eds), p-adic Analysis. Lecture Notes in Mathematics, vol. 1454, pp. 194 - 204, Springer, Berlin, Heidelberg.
• Matthijs J. Coster, Supercongruences.

Crossrefs

Cf. A112029, A176331, A176332, A176334.

Programs

GAP

T:= function(n,k)
    return Sum([0..n], j-> (-1)^(n-j)*Binomial(j,k)*Binomial(j,n-k) );
  end;
List([0..30], n-> T(2*n,n) ); # G. C. Greubel, Dec 07 2019

Magma

T:= func< n,k | &+[(-1)^(n-j)*Binomial(j,n-k)*Binomial(j,k): j in [0..n]] >;
[T(2*n,n): n in [0..30]]; // G. C. Greubel, Dec 07 2019

Maple

A176335 := proc(n)
    add((-1)^k*binomial(k,n)^2,k=0..2*n);
end proc: # R. J. Mathar, Feb 10 2015

Mathematica

T[n_, k_]:= Sum[(-1)^(n-j)*Binomial[j, k]*Binomial[j, n-k], {j,0,n}]; Table[T[2*n, n], {n,0,30}] (* G. C. Greubel, Dec 07 2019 *)

PARI

T(n,k) = sum(j=0, n, (-1)^(n-j)*binomial(j, n-k)*binomial(j, k));
vector(31, n, T(2*(n-1), n-1) ) \\ G. C. Greubel, Dec 07 2019

Sage

@CachedFunction
def T(n, k): return sum( (-1)^(n-j)*binomial(j, n-k)*binomial(j, k) for j in (0..n))
[T(2*n, n) for n in (0..30)] # G. C. Greubel, Dec 07 2019

Formula

a(n) = Sum_{k=0..2n} C(k,n)^2*(-1)^k.

Conjecture: 224*n^2*(n-1)*a(n) - 48*(n-1)*(65*n^2 - 36*n - 13)*a(n-1) + 4*(-1839*n^3 + 11081*n^2 - 21932*n + 14280)*a(n-2) + 12*(-81*n^3 + 326*n^2 - 591*n + 562)*a(n-3) - (n-3)*(1853*n^2 - 7403*n + 7140)*a(n-4) - 12*(n-4)*(2*n-7)^2*a(n-5) = 0. - R. J. Mathar, Feb 10 2015

From Peter Bala, Aug 08 2024: (Start)

a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+k, k)^2. Cf. A112029.

Conjecture (assuming an offset of 1): the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 5 and all positive integers n and r [added Nov 29 2024: proved by Coster. See Theorem 4]. (End)

a(n) ~ 2^(4*n+2) / (5*Pi*n). - Vaclav Kotesovec, Aug 08 2024

a(n) = binomial(2*n, n)^2 * hypergeom([1, -n, -n], [-2*n, -2*n], -1). - Peter Luschny, Nov 29 2024