A176735 a(n) = (n+8)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.

1, 9, 91, 1019, 12501, 166589, 2394751, 36920799, 607496041, 10622799089, 196677847971, 3843107102339, 79025598374461, 1705654851091749, 38551739502886471, 910569176481673319, 22431936328103456721, 575367515026293191129, 15340898308261381733611, 424560869593530584247819

Offset: 0

Author

Wolfdieter Lang, Jul 14 2010

Keywords

• nonn
• easy

Comments

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=9 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A049389(n) = (n+8)!/8!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs 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).

Examples

			Necklaces and 9 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 !4*1,binomial(4,3)*!3*c9(1), (binomial(4,2)*! 2)*c9(2), and 1*c9(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c9(n):=A049389(n) numbers for the pure 9-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=9: 1/(1-x)^9). This adds up as 9 + 4*2*9 + (6*1)*90 + 11880 = 12501 = a(4).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A176734 (necklaces and k=8 cords).

Programs

Mathematica

RecurrenceTable[{a[0]==1,a[1]==9,a[n]==(n+8)a[n-1]+(n-1)a[n-2]},a[n],{n,20}] (* Harvey P. Dale, Oct 20 2011 *)
Table[(-1)^n HypergeometricPFQ[{10, -n}, {}, 1], {n, 0, 20}] (* Benedict W. J. Irwin, May 27 2016 *)

Formula

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^9) = exp(-x)/(1-x)^10, equivalent to the given recurrence.

a(n) = A086764(n+9,9).

a(n) = (-1)^n*2F0(10,-n;;1). - Benedict W. J. Irwin, May 27 2016