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%I #12 Apr 25 2015 19:47:42
%S 1,6,43,356,3333,34754,398959,4996032,67741129,988344062,15434831091,
%T 256840738076,4536075689293,84731451264186,1668866557980343,
%U 34563571477305464,750867999393119889,17072113130285524982,405423357986250112699,10037458628015142154452,258639509502117306002581
%N a(n) = (n+5)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.
%C 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=6 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 {A001725(n+5) = (n+5)!/5!}. 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).
%F E.g.f. (exp(-x)/(1-x))*(1/(1-x)^6) = exp(-x)/(1-x)^7, equivalent to the recurrence.
%F a(n) = A086764(n+6,6).
%F a(n) = A090010(n), n>0. - _R. J. Mathar_, Jul 22 2010
%F a(n) = (-1)^n*hypergeom([-n,7],[],1). - _Peter Luschny_, Apr 25 2015
%e Necklaces and 6 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*c6(1), (binomial(4,2)*2)*c6(2), and 1*c6(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c6(n):=A001725(n+5) numbers for the pure 6-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=6: 1/(1-x)^6). This adds up as 9 + 4*2*6 + (6*1)*42 + 3024 = 3333 = a(4).
%p a := n -> hypergeom([-n,7],[],1)*(-1)^n:
%p seq(simplify(a(n)),n=0..9); # _Peter Luschny_, Apr 25 2015
%t Rest[RecurrenceTable[{a[0]==1,a[-1]==0,a[n]==(n+5)a[n-1]+(n-1)a[n-2]},a,{n,20}]] (* _Harvey P. Dale_, Oct 01 2012 *)
%Y Cf. A000153, A000261, A001909, A001910 (necklaces and k=5 cords), A176732.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Jul 14 2010
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