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A005990
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(n-1)*(n+1)!/6.
(Formerly M4551)
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14
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0, 1, 8, 60, 480, 4200, 40320, 423360, 4838400, 59875200, 798336000, 11416204800, 174356582400, 2833294464000, 48819843072000, 889218570240000, 17072996548608000, 344661117825024000, 7298706024529920000, 161787983543746560000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Coefficients of Gandhi polynomials.
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
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]
Contribution from Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 26 2009: (Start)
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.
(End)
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REFERENCES
| D. Dumont, Interpretations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 1..300
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
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FORMULA
| a(n) = A052571(n+2)/6. - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 11 2007
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
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]
a(n) = (n+3)!/2 * sum((k+1)!/(k+3)!,k=1..n), with offset 0 [From Gary Detlefs (gdetlefs(AT)aol.com), Aug 05 2010]
a(n) = (n+2)!*sum(1/((2*k+4)*(k+3)),k=1..n-1). [From Gary Detlefs, Oct 09 2011]
a(n) = (n+2)!*(1+3*(H(n+1)-H(n+2)))/6, where H(n) is the n-th harmonic number. [From Gary Detlefs, Oct 09 2011]
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EXAMPLE
| E.g.f.: (-1+3*x)/(3!*(1-x)^3), a(0)=-1/3!. Such e.g.f. computations resulted from e-mail exchange with Gary Detlefs. [From Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), May 27 2010]
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MAPLE
| [ seq((n-1)*(n+1)!/6, n=1..40) ];
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
seq(sum(mul(j, j=3..n), k=3..n)/3, n=2..21); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 01 2007
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]
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MATHEMATICA
| Table[Sum[n!/6, {i, 3, n}], {n, 2, 21}] [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 12 2009]
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PROG
| (MAGMA) [(n-1)*Factorial(n+1)/6: n in [1..25]]; // Vincenzo Librandi, Oct 11 2011
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CROSSREFS
| A090672(n)/2.
Cf. A001715.
Equals the second right hand column of A167568 divided by 2. - Johannes W. Meijer (meijgia(AT)hotmail.com), Nov 12 2009
Sequence in context: A099156 A199526 A129331 * A160228 A099337 A075147
Adjacent sequences: A005987 A005988 A005989 * A005991 A005992 A005993
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Formula from Robert Newstedt.
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