

A176733


a(n) = (n+6)*a(n1) + (n1)*a(n2), a(1)=0, a(0)=1.


6



1, 7, 57, 527, 5441, 61959, 770713, 10391023, 150869313, 2346167879, 38896509881, 684702346767, 12752503850497, 250514001320647, 5176062576469401, 112204510124346479, 2546140161382663553, 60356495873790805383, 1491840283714484609593, 38382424018590349736719
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OFFSET

0,2


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=7 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 {A001730(n+6) = (n+6)!/6!}. 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).


LINKS



FORMULA

E.g.f. (exp(x)/(1x))*(1/(1x)^7) = exp(x)/(1x)^8, equivalent to the given recurrence.


EXAMPLE

Necklaces and 7 cords problem. For n=4 one considers the following weak 2part 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*c7(1), (binomial(4,2)*!2)*c7(2), and 1*c7(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c7(n):=A001730(n+6) numbers for the pure 7cord problem (see the remark on the e.g.f. for the kcord problem in A000153; here for k=7: 1/(1x)^7). This adds up as 9 + 4*2*7 + (6*1)*56 + 5040 = 5441 = a(4).


MATHEMATICA



CROSSREFS

Cf. A176732 (necklaces and k=6 cords).


KEYWORD

nonn,easy


AUTHOR



STATUS

approved



