A343832 a(n) = Sum_{k=0..n} k! * binomial(n,k) * binomial(2*n+1,k).

1, 4, 31, 358, 5509, 106096, 2456299, 66471826, 2059640713, 71920704124, 2794938616471, 119653108240414, 5595650767265101, 283841520215780008, 15523069639558351459, 910529206043204428426, 57023540590242398853649, 3797750659849704886903156, 268025698704886063968108943

Offset: 0

Author

Seiichi Manyama, May 01 2021

Keywords

• nonn

Comments

Let A(x) be the e.g.f. of this sequence, and B(x) be the e.g.f. of A082545, then A(x)/B(x) = C(x) where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108). This follows from the fact that this sequence and A082545 form adjacent semi-diagonals of table A088699. - Paul D. Hanna, Aug 16 2022

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..365
• Wikipedia, Laguerre polynomials
• Index entries for sequences related to Laguerre polynomials

Crossrefs

Cf. A000522, A045721, A082545, A152059, A248668, A343830, A343831, A000108.

Programs

Magma

[Factorial(n)*Evaluate(LaguerrePolynomial(n, n+1), -1): n in [0..40]]; // G. C. Greubel, Aug 11 2022

Maple

a := n -> add(k!*binomial(n, k)*binomial(2*n+1, k), k=0..n):
a := n -> n!*add(binomial(2*n+1, k)/(n-k)!, k=0..n):
a := n -> (-1)^n*KummerU(-n, n+2, -1):
a := n -> n!*LaguerreL(n, n+1, -1): # Peter Luschny, May 02 2021

Mathematica

a[n_] := Sum[k! * Binomial[n, k] * Binomial[2*n+1, k], {k, 0, n}]; Array[a, 20, 0] (* Amiram Eldar, May 01 2021 *)
Table[(-1)^n * HypergeometricU[-n, 2 + n, -1], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2021 *)

PARI

a(n) = sum(k=0, n, k!*binomial(n, k)*binomial(2*n+1, k));

PARI

a(n) = (2*n+1)!*sum(k=0, n, binomial(n, k)/(k+n+1)!);

PARI

a(n) = n!*sum(k=0, n, binomial(2*n+1, k)/(n-k)!);

PARI

a(n) = n!*pollaguerre(n, n+1, -1);

SageMath

[factorial(n)*gen_laguerre(n, n+1, -1) for n in (0..40)] # G. C. Greubel, Aug 11 2022

Formula

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

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

a(n) = n! * LaguerreL(n, n+1, -1).

a(n) = n! * [x^n] exp(x/(1 - x))/(1 - x)^(n+2).

a(n) == 1 (mod 3).

a(n) ~ 2^(2*n + 3/2) * n^n / exp(n-1). - Vaclav Kotesovec, May 02 2021

From Paul D. Hanna, Aug 16 2022: (Start)

E.g.f.: exp( (1-2*x - sqrt(1-4*x))/(2*x) ) * (1 - sqrt(1-4*x)) / (2*x*sqrt(1-4*x)), derived from the e.g.f for A082545 given by Mark van Hoeij.

E.g.f.: exp(C(x) - 1) * C(x) / sqrt(1-4*x), where C(x) = (1 - sqrt(1-4*x))/(2*x) is the Catalan function (A000108). (End)