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A288312
Number of endofunctions on [2n] such that the image size equals n.
4
1, 2, 84, 10800, 2857680, 1285956000, 880599202560, 853262368358400, 1111400775560275200, 1873276460474747328000, 3967400888465895264384000, 10313998054713896966296473600, 32291970618091110826769565696000, 119851615755915509174015455948800000
OFFSET
0,2
LINKS
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
KEYWORD
nonn
AUTHOR
Alois P. Heinz, Jun 07 2017
STATUS
approved