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A000153
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a(n) = n*a(n-1) + (n-2)*a(n-2), with a(0) = 0, a(1) = 1.
(Formerly M1791 N0706)
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32
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0, 1, 2, 7, 32, 181, 1214, 9403, 82508, 808393, 8743994, 103459471, 1328953592, 18414450877, 273749755382, 4345634192131, 73362643649444, 1312349454922513, 24796092486996338, 493435697986613143, 10315043624498196944
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OFFSET
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0,3
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COMMENTS
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With offset 1, permanent of (0,1)-matrix of size n X (n+d) with d=2 and n zeros not on a line. This is a special case of Theorem 2.3 of Seok-Zun Song et al. Extremes of permanents of (0,1)-matrices, pp. 201-202. - Jaap Spies, Dec 12 2003
Starting (1, 2, 7, 32, ...) = inverse binomial transform of A001710 starting (1, 3, 12, 60, 360, 2520, ...). - Gary W. Adamson, Dec 25 2008
This sequence appears in Euler's analysis of the divergent series 1 - 1! + 2! - 3! + 4! ..., see Sandifer. For information about this and related divergent series see A163940. - Johannes W. Meijer, Oct 16 2009
a(n+1)=:b(n), n>=1, enumerates the ways to distribute n beads labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and two indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as a beadless cords contribute each a factor 1 in the counting, e.g., b(0):= 1*1 = 1. See A000255 for the description of a fixed cord with beads.
This produces for b(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and {(n+1)!}={A000042(n+1}. This follows from the general problem with only k indistinguishable, ordered, fixed cords which has e.g.f. 1/(1-x)^k, and the pure necklace problem (no necklaces with one bead allowed) with e.g.f. for the subfactorials. Therefore also the recurrence b(n) = (n+1)*b(n-1) + (n-1)*b(n-2) with b(-1)=0 and b(0)=1 holds.
This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010
a(n) is a function of the subfactorials..sf... A000166(n) a(n) = (n*sf(n+1) - (n+1)*sf(n))/(2*n*(n-1)*(n+1)),n>1, with offset 1. - Gary Detlefs, Nov 06 2010
For even k the sequence a(n) (mod k) is purely periodic with exact period a divisor of k, while for odd k the sequence a(n) (mod k) is purely periodic with exact period a divisor of 2*k. See A047974. - Peter Bala, Dec 04 2017
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REFERENCES
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Brualdi, Richard A. and Ryser, Herbert J., Combinatorial Matrix Theory, Cambridge NY (1991), Chapter 7.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 188.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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E.g.f.: ( 1 - x )^(-3)*exp(-x), for offset 1.
a(n) = round(1/2*(n^2 + 3*n + 1)*n!/exp(1))/n , n>=1. - Simon Plouffe, March 1993
G.f.: hypergeom([1,3],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
G.f.: (1+x)^2/(2*x*Q(0)) - 1/(2*x) - 1, where Q(k) = 1 - 2*k*x - x^2*(k + 1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 08 2013
G.f.: -1/G(0), where G(k) = 1 + 1/(1 - (1+x)/(1 + x*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 01 2013
G.f.: x/Q(0), where Q(k) = 1 - 2*x*(k+1) - x^2*(k+1)*(k+3)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2013
a(n) = hypergeom([3, -n+1], [], 1))*(-1)^(n+1) for n>=1. - Peter Luschny, Sep 20 2014
a(n) = KummerU(-n + 1, -n - 1, -1) for n >= 1. - Peter Luschny, May 10 2022
a(n) = (n^2 + 3*n + 1)*Gamma(n,-1)/(2*exp(1)) + (1 + n/2)*(-1)^n for n >= 1. - Martin Clever, Apr 06 2023
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EXAMPLE
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Necklaces and 2 cords problem. For n=4 one considers the following weak 2 part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively sf(4)*1,binomial(4,3)*sf(3)*c2(1), (binomial(4,2)*sf(2))*c2(2), and 1*c2(4) with the subfactorials sf(n):=A000166(n) (see the necklace comment there) and the c2(n):=(n+1)! numbers for the pure 2 cord problem (see the above given remark on the e.g.f. for the k cords problem; here for k=2: 1/(1-x)^2). This adds up as 9 + 4*2*2 + (6*1)*6 + 120 = 181 = b(4) = A000153(5). - Wolfdieter Lang, Jun 02 2010
G.f. = x + 2*x^2 + 7*x^3 + 32*x^4 + 181*x^5 + 1214*x^6 + 9403*x^7 + 82508*x^8 + ...
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MAPLE
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f:= n-> floor(((n+1)!+1)/e): g:=n-> (n*f(n+1)-(n+1)*f(n))/(2*n*(n-1)*(n+1)):seq( g(n), n=2..20); # Gary Detlefs, Nov 06 2010
a := n -> `if`(n=0, 0, hypergeom([3, -n+1], [], 1))*(-1)^(n+1); seq(simplify(a(n)), n=0..20); # Peter Luschny, Sep 20 2014
0, seq(simplify(KummerU(-n + 1, -n - 1, -1)), n = 1..20); # Peter Luschny, May 10 2022
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MATHEMATICA
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nn = 20; Prepend[Range[0, nn]!CoefficientList[Series[Exp[-x]/(1 - x)^3, {x, 0, nn}], x], 0] (* Geoffrey Critzer, Oct 28 2012 *)
RecurrenceTable[{a[0]==0, a[1]==1, a[n]==n a[n-1]+(n-2)a[n-2]}, a, {n, 20}] (* Harvey P. Dale, May 08 2013 *)
a[ n_] := If[ n < 1, 0, (n - 1)! SeriesCoefficient[ Exp[ -x] / (1 - x)^3, {x, 0, n - 1}]]; (* Michael Somos, Jun 01 2013 *)
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1, 3}, {}, x / (x + 1)] x / (x + 1), {x, 0, n}]; (* Michael Somos, Jun 01 2013 *)
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PROG
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(Haskell)
a000153 n = a000153_list !! n
a000153_list = 0 : 1 : zipWith (+)
(zipWith (*) [0..] a000153_list) (zipWith (*) [2..] $ tail a000153_list)
(PARI) x='x+O('x^66); concat([0], Vec(x*serlaplace(exp(-x)/(1-x)^3))) \\ Joerg Arndt, May 08 2013
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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