0,2

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=10 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 {A049398(n) = (n+9)!/9!}. 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).

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^10) = exp(-x)/(1-x)^11, equivalent to the given recurrence.

a(n) = A086764(n+10,10).

Necklaces and 10 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*c10(1), (binomial(4,2)*! 2)*c10(2), and 1*c10(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c10(n):=A049398(n) numbers for the pure 10-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=10: 1/(1-x)^10). This adds up as 9 + 4*2*10 + (6*1)*110 + 17160 = 17909 = a(4).

Cf. A176735 (necklaces and k=9 cords).

Sequence in context: A014993 A015592 A122574 * A084031 A210507 A066275

Adjacent sequences: A176733 A176734 A176735 * A176737 A176738 A176739

nonn,easy

Wolfdieter Lang, Jul 14 2010

approved