Search: id:A055775
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%I A055775
%S A055775 1,1,2,4,10,26,64,163,416,1067,2755,7147,18613,48638,127463,334864,
%T A055775 881657,2325750,6145596,16263866,43099804,114356611,303761260,
%U A055775 807692034,2149632061,5726042115,15264691107,40722913454,108713644516
%N A055775 Floor(n^n/n!).
%C A055775 Stirling's approximation for n! suggests that this should be about e^n/
               sqrt(pi*2n). R. W. Gosper has noted that e^n/sqrt(pi*(2n+1/3)) is 
               significantly better.
%C A055775 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
%H A055775 Leroy Quet, <a href="http://www.prism-of-spirals.net/">Home Page</a> 
               (listed in lieu of email address)
%H A055775 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               StirlingsApproximation.html">Stirling's Approximation for n!</a>
%F A055775 a(n) =[A000312(n)/A000142(n)]
%e A055775 a(5)=26 since 5^5=3125, 5!=120, 3125/120=26.0416666...
%t A055775 lst={}; Do[AppendTo[lst,Floor[n^n/n! ]],{n,5!}]; lst [From Vladimir Orlovsky 
               (4vladimir(AT)gmail.com), Jan 15 2009]
%Y A055775 Cf. A073225, A094082.
%Y A055775 Cf. A053042, A036679, A061711 [From Vladimir Orlovsky (4vladimir(AT)gmail.com), 
               Jan 15 2009]
%Y A055775 Sequence in context: A100605 A154322 A090031 this_sequence A090032 A090377 
               A151278
%Y A055775 Adjacent sequences: A055772 A055773 A055774 this_sequence A055776 A055777 
               A055778
%K A055775 nonn
%O A055775 0,3
%A A055775 Henry Bottomley (se16(AT)btinternet.com), Jul 12 2000
%E A055775 More terms from James A. Sellers (sellersj(AT)math.psu.edu), Jul 13 2000

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