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

Table[(-1)^n HypergeometricPFQ[{10, -n}, {}, 1], {n, 0, 20}] (* Benedict W. J. Irwin, May 27 2016 *)

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

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a(n) enumerates the possibilities to distribute 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 a beadless cords contributes 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 ( Febr Feb 27 2010).

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 cords -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).

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RecurrenceTable[{a[0]==1, a[1]==9, a[n]==(n+8)a[n-1]+(n-1)a[n-2]}, a[n], {n, 20}] (* From _Harvey P. Dale, _, Oct 20 2011 *)