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%I #30 Sep 08 2022 08:45:01
%S 1,1,2,4,10,26,64,163,416,1067,2755,7147,18613,48638,127463,334864,
%T 881657,2325750,6145596,16263866,43099804,114356611,303761260,
%U 807692034,2149632061,5726042115,15264691107,40722913454,108713644516
%N a(n) = floor(n^n / n!).
%C 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.
%C 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
%C 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
%H Vincenzo Librandi, <a href="/A055775/b055775.txt">Table of n, a(n) for n = 0..300</a>
%H Washington Bomfim, <a href="/A208846/a208846.txt">A method to find bijections from a set of n integers to {0,1, ... ,n-1}</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StirlingsApproximation.html">Stirling's Approximation for n!</a>
%F a(n) = floor(A000312(n)/A000142(n)).
%e a(5)=26 since 5^5=3125, 5!=120, 3125/120=26.0416666...
%t Join[{1}, Table[Floor[n^n/n!], {n, 30}]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 15 2009 *)
%o (Magma) [Floor((n^n)/Factorial(n)): n in [0..30]]; // _Vincenzo Librandi_, Aug 22 2011
%o (PARI) a(n)=n^n\n! \\ _Charles R Greathouse IV_, Apr 17 2012
%Y Cf. A073225, A094082, A053042, A036679, A061711, A152170, A209081, A208846, A208847.
%K nonn,easy
%O 0,3
%A _Henry Bottomley_, Jul 12 2000
%E More terms from _James A. Sellers_, Jul 13 2000
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