A055775 - OEIS

A055775

a(n) = floor(n^n / n!).

1, 1, 2, 4, 10, 26, 64, 163, 416, 1067, 2755, 7147, 18613, 48638, 127463, 334864, 881657, 2325750, 6145596, 16263866, 43099804, 114356611, 303761260, 807692034, 2149632061, 5726042115, 15264691107, 40722913454, 108713644516

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OFFSET

0,3

COMMENTS

Stirling's approximation for n! suggests that this should be about e^n/sqrt(pi*2n). Bill Gosper has noted that e^n/sqrt(pi*(2n+1/3)) is significantly better.

n^n/n! = A001142(n)/A001142(n-1), where A001142(n) is product{k=0 to n} C(n,k) (where C() is a binomial coefficient). - Leroy Quet, May 01 2004

There are n^n distinct functions from [n] to [n] or sequences on n symbols of length n, the number of those sequences having n distinct symbols is n!. So the probability P(n) of bijection is n!/n^n. The expected value of the number of functions that we pick until we found a bijection is the reciprocal of P(n), or n^n/n!. - Washington Bomfim, Mar 05 2012

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Washington Bomfim, A method to find bijections from a set of n integers to {0,1, ... ,n-1}

Eric Weisstein's World of Mathematics, Stirling's Approximation for n!

FORMULA

a(n) = floor(A000312(n)/A000142(n)).

EXAMPLE

a(5)=26 since 5^5=3125, 5!=120, 3125/120=26.0416666...

MATHEMATICA

Join[{1}, Table[Floor[n^n/n!], {n, 30}]] (* Vladimir Joseph Stephan Orlovsky, Jan 15 2009 *)

PROG

(Magma) [Floor((n^n)/Factorial(n)): n in [0..30]]; // Vincenzo Librandi, Aug 22 2011