A288312 Number of endofunctions on [2n] such that the image size equals n.

1, 2, 84, 10800, 2857680, 1285956000, 880599202560, 853262368358400, 1111400775560275200, 1873276460474747328000, 3967400888465895264384000, 10313998054713896966296473600, 32291970618091110826769565696000, 119851615755915509174015455948800000

Offset

0,2

Links

Alois P. Heinz, Table of n, a(n) for n = 0..195

Formula

a(n) = Stirling2(2*n,n) * n! * binomial(2*n,n).

a(n) = A090657(2n,n) = A101817(2n,n) = A219859(2n,n).

a(n) ~ n^(2*n - 1/2) * 2^(4*n) / (sqrt(Pi*(1-c)) * c^n * (2-c)^n * exp(2*n)), where c = -LambertW(-2*exp(-2)) = -A226775 = 0.4063757399599599... - Vaclav Kotesovec, Jun 10 2017

Example

a(1) = 2: (1,1), (2,2).

Maple

b:= proc(n, k) option remember; `if`(k=n, n!,
      `if`(k=0, 0, n*(b(n-1, k-1)+b(n-1, k)*k/(n-k))))
    end:
a:= n-> b(2*n, n):
seq(a(n), n=0..15);

Mathematica

Table[StirlingS2[2*n, n]*(2*n)!/n!, {n, 0, 20}] (* Vaclav Kotesovec, Jun 10 2017 *)

Prog

(PARI) a(n)=stirling(2*n, n, 2)*n!*binomial(2*n, n); \\ Indranil Ghosh, Jul 04 2017
(Python)
from mpmath import *
mp.dps=100
def a(n): return int(stirling2(2*n, n)*fac(n)*binomial(2*n, n))
print([a(n) for n in range(21)]) # Indranil Ghosh, Jul 04 2017

Crossrefs

Cf. A000142, A048993, A090657, A101817, A219859.

Sequence in context: A265591 A348265 A157063 * A289198 A318128 A181119

Adjacent sequences: A288309 A288310 A288311 * A288313 A288314 A288315

Keyword

nonn

Author

Alois P. Heinz, Jun 07 2017

Status

approved