

A055775


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


25



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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(n1), 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, ... ,n1}
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
(PARI) a(n)=n^n\n! \\ Charles R Greathouse IV, Apr 17 2012


CROSSREFS

Cf. A073225, A094082, A053042, A036679, A061711, A152170, A209081, A208846, A208847.
Sequence in context: A183947 A154322 A090031 * A239076 A217988 A295704
Adjacent sequences: A055772 A055773 A055774 * A055776 A055777 A055778


KEYWORD

nonn,easy


AUTHOR

Henry Bottomley, Jul 12 2000


EXTENSIONS

More terms from James A. Sellers, Jul 13 2000


STATUS

approved



