|
| |
|
|
A001909
|
|
a(n) = n*a(n-1) + (n-4)*a(n-2), a(2) = 0, a(3) = 1.
(Formerly M3576 N1450)
|
|
24
|
|
|
|
0, 1, 4, 21, 134, 1001, 8544, 81901, 870274, 10146321, 128718044, 1764651461, 25992300894, 409295679481, 6860638482424, 121951698034461, 2291179503374234, 45361686034627361, 943892592746534964, 20592893110265899381, 470033715095287415734
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
2,3
|
|
|
COMMENTS
|
With offset 1, permanent of (0,1)-matrix of size n X (n+d) with d=4 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
a(n+3)=: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 four 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 the sequence {A001715 (n+3)}. See the necklaces and cords problem comment in A000153. Therefore also the recurrence b(n) = (n+3)*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
|
|
|
REFERENCES
|
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).
|
|
|
LINKS
|
|
|
|
FORMULA
|
E.g.f.: exp(-x) / (1 - x)^5 = Sum_{k>=0} a(k+3) * x^k / k!. - Michael Somos, Feb 19 2003
G.f.: x*hypergeom([1,5],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
a(n) = hypergeom([5,-n+3],[],1))*(-1)^(n+1) for n>=3. - Peter Luschny, Sep 20 2014
|
|
|
EXAMPLE
|
Necklaces and four 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)*c4(1), (binomial(4,2)*sf(2))*c4(2), and 1*c4(4) with the subfactorials sf(n):=A000166(n) (see the necklace comment there) and the c4(n):=A001715(n+3) = (n+3)!/3! numbers for the pure 4 cord problem (see the remark on the e.g.f. for the k cords problem in A000153; here for k=4: 1/(1-x)^4). This adds up as 9 + 4*2*4 + (6*1)*20 + 840 = 1001 = b(4) = A001909(7). - Wolfdieter Lang, Jun 02 2010
x^3 + 4*x^4 + 21*x^5 + 134*x^6 + 1001*x^7 + 8544*x^8 + 81901*x^9 + 870274*x^10 + ...
|
|
|
MAPLE
|
a := n -> `if`(n<4, n-2, hypergeom([5, -n+3], [], 1))*(-1)^(n+1);
seq(round(evalf(a(n), 100)), n=2..22); # Peter Luschny, Sep 20 2014
|
|
|
MATHEMATICA
|
t = {0, 1}; Do[AppendTo[t, n*t[[-1]] + (n-4)*t[[-2]]], {n, 4, 20}]; t (* T. D. Noe, Aug 17 2012 *)
nxt[{n_, a_, b_}]:={n+1, b, b(n+1)+a(n-3)}; NestList[nxt, {3, 0, 1}, 20][[All, 2]] (* Harvey P. Dale, Jul 17 2018 *)
|
|
|
PROG
|
(PARI) {a(n) = if( n<2, 0, -contfracpnqn( matrix(2, n, i, j, j - 4*(i==1))) [1, 1])} /* Michael Somos, Feb 19 2003 */
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
