OFFSET
0,4
COMMENTS
a(n) enumerates circular permutations of {1,2,...,n+9} with exactly nine successor pairs (i,i+1). Due to cyclicity also (n+9,1) is a successor pair.
REFERENCES
Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 183, eq. (5.15), for k=9.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..440
FORMULA
a(n) = binomial(n+9,9)*A000757(n), n>=0.
E.g.f.: (d^9/dx^9) (x^9/9!)*(1-log(1-x))/e^x.
EXAMPLE
a(0)=1 because from the 9!/9 = 40320 circular permutations of n=9 elements only one, namely (1,2,3,4,5,6,7,8,9), has nine successors.
MATHEMATICA
f[n_] := (-1)^n + Sum[(-1)^k*n!/((n - k)*k!), {k, 0, n - 1}]; a[n_, n_] = 1; a[n_, 0] := f[n]; a[n_, k_] := a[n, k] = n/k*a[n - 1, k - 1]; Table[a[n, 9], {n, 9, 25}] (* G. C. Greubel, Nov 10 2016 *)
PROG
(PARI) a(n)=((-1)^n + sum( k=0, n-1, (-1)^k * binomial( n, k) * (n - k - 1)!))*binomial(n+9, 9) \\ Charles R Greathouse IV, Nov 10 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Jan 21 2008, Feb 22 2008
STATUS
approved