Search: id:A005990
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%I A005990 M4551
%S A005990 0,1,8,60,480,4200,40320,423360,4838400,59875200,798336000,11416204800,
%T A005990 174356582400,2833294464000,48819843072000,889218570240000,17072996548608000,
%U A005990 344661117825024000,7298706024529920000,161787983543746560000
%N A005990 (n-1)*(n+1)!/6.
%C A005990 Coefficients of Gandhi polynomials.
%C A005990 a(n) = Sum_{pi in Symm(n)} Sum_{i=1..n} max(pi(i)-i,0), i.e. the total 
               positive displacement of all letters in all permutations on n letters. 
               - Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Oct 25 2006
%C A005990 a(n) is also the sum of the excedances of all permutations of [n]. An 
               excedance of a permutation p of [n] is an i (1<=i<=n-1) such that 
               p(i)>i. Proof: i is an excedance if p(i)=i+1, i+2, ..., n (n-i possibilities), 
               with the remaining values of p forming any permutation of [n]\{p(i)} 
               in the positions [n]\{i} ((n-1)! possibilities). Summation of i(n-i)(n-1)! 
               over i from 1 to n-1 completes the proof. Example: a(3)=8 because 
               the permutations 123, 132, 213, 231, 312, 321 have excedances NONE, 
               {2}, {1}, {1,2}, {1}, {1}, respectively. [From Emeric Deutsch (deutsch(AT)duke.poly.edu), 
               Oct 26 2008]
%C A005990 Contribution from Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 26 2009: 
               (Start)
%C A005990 a(n) is also the number of doubledescents in all permutations of {1,2,
               ...,n-1}. We say that i is a doubledescent of a permutation p if 
               p(i)>p(i+1)>p(i+2). Example: a(3)=8 because each of the permutations 
               1432, 4312, 4213, 2431, 3214, 3421 has one doubledescent, the permutation 
               4321 has two doubledescents and the remaining 17 permutations of 
               {1,2,3,4} have no doubledescents.
%C A005990 (End)
%D A005990 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A005990 D. Dumont, Interpretations combinatoires des nombres de Genocchi, Duke 
               Math. J., 41 (1974), 305-318.
%H A005990 Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Enumerative Formulas 
               for Some Functions on Finite Sets</a>
%F A005990 a(n)=A052571(n+2)/6. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               May 11 2007
%F A005990 a(n)=sum(sum(sum(n!/6, j=1..n),k=-1..n),m=0..n), n>=0 . - Zerinvary Lajos 
               (zerinvarylajos(AT)yahoo.com), May 11 2007
%F A005990 If we define f(n,i,x)= sum(sum(binomial(k,j)*stirling1(n,k)*stirling2(j,
               i)*x^(k-j),j=i..k),k=i..n) then a(n+1)=(-1)^(n-1)*f(n,1,-4), (n>=1). 
               [From Milan R. Janjic (agnus(AT)blic.net), Mar 01 2009]
%p A005990 [ seq((n-1)*(n+1)!/6,n=1..40) ];
%p A005990 a:=n->sum(sum(sum(n!/6, j=1..n),k=-1..n),m=0..n): seq(a(n), n=0..19); 
               - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 11 2007
%p A005990 seq(sum(mul(j,j=3..n), k=3..n)/3, n=2..21); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Jun 01 2007
%p A005990 restart: G(x):=x^3/(1-x)^2: f[0]:=G(x): for n from 1 to 21 do f[n]:=diff(f[n-1],
               x) od: x:=0: seq(f[n]/3!,n=2..21);# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Apr 01 2009]
%t A005990 Table[Sum[n!/6, {i, 3, n}], {n, 2, 21}] [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Jul 12 2009]
%Y A005990 A090672(n)/2.
%Y A005990 Cf. A001715.
%Y A005990 Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Nov 12 
               2009: (Start)
%Y A005990 Equals the second right hand column of A167568 divided by 2.
%Y A005990 (End)
%Y A005990 Sequence in context: A001267 A099156 A129331 this_sequence A160228 A099337 
               A075147
%Y A005990 Adjacent sequences: A005987 A005988 A005989 this_sequence A005991 A005992 
               A005993
%K A005990 nonn,easy
%O A005990 1,3
%A A005990 N. J. A. Sloane (njas(AT)research.att.com).
%E A005990 Formula from Robert Newstedt.

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