A241023 Central terms of the triangle in A102413.

1, 4, 16, 76, 384, 2004, 10672, 57628, 314368, 1728292, 9560016, 53144172, 296642688, 1661529588, 9333781872, 52566230076, 296697618432, 1677889961028, 9505151782288, 53928746972812, 306393243712384, 1742920028025364, 9925790375394096, 56584659163097436

Offset: 0

Author

Reinhard Zumkeller, Apr 15 2014

Keywords

• nonn
• easy

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A001850, A006318, A047781, A102413, A333481.

Programs

Haskell

a241023 n = a102413 (2 * n) n

Mathematica

a[0] = 1; a[n_] := 4 Hypergeometric2F1[1 - n, n + 1, 1, -1];
Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Jun 28 2019 *)

Formula

a(n) = 4 * A047781(n).

a(n) = A102413(2*n,n).

a(n) = 2*Hyper2F1([-n, n], [1], -1) for n>0. - Peter Luschny, Aug 02 2014

D-finite g.f. = (1+x)/sqrt(1-6*x+x^2), pairwise sums of A001850. - R. J. Mathar, Jan 15 2020

From Peter Bala, Apr 16 2024: (Start)

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

a(n) = (-1)^(n+1) * 4*n * hypergeom([n+1, -n+1], [2], 2).

n*(2*n - 3)*a(n) = 4*(3*n^2 - 6*n + 2)*a(n-1) - (2*n - 1)*(n - 2)*a(n-2) with a(0) = 1 and a(1) = 4.

O.g.f.: Sum_{n >= 0} (2^n)*binomial(2*n,n)*x^n/(1 + x)^(2*n) = 1 + 4*x + 16*x^2 + 76*x^3 + 384*x^4 + .... (End)

From Peter Bala, Sep 18 2024: (Start)

a(n) = [x^n] 1/S(-x)^(2*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the large Schröder numbers A006318. Cf. A333481.

The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.

Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r. (End)