|
|
A135802
|
|
Fifth column (k=4) of triangle A134832 (circular succession numbers).
|
|
3
|
|
|
1, 0, 0, 35, 70, 1008, 7560, 75570, 804375, 9443720, 120408288, 1658028645, 24515212540, 387332966720, 6511826843280, 116059273664436, 2185693176650685, 43366955622595920, 904164368153680480
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,4
|
|
COMMENTS
|
a(n) enumerates circular permutations of {1,2,...,n+4} with exactly four successor pairs (i,i+1). Due to cyclicity also (n+4,1) is a successor pair.
|
|
REFERENCES
|
Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 183, eq. (5.15), for k=4.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = binomial(n+4,4)*A000757(n), n>=0.
E.g.f.: (d^4/dx^4) (x^4/4!)*(1-log(1-x))/e^x.
|
|
EXAMPLE
|
a(0)=1 because the 4!/4 = 6 circular permutations of n=4 elements (1,2,3,4), (1,4,3,2), (1,3,4,2),(1,2,4,3), (1,4,2,3) and (1,3,2,4) have 4,0,1,1,1 and 1 successor pair, respectively. Hence (1,2,3,4) is the only circular permutation with 4 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, 4], {n, 4, 25}] (* G. C. Greubel, Nov 10 2016 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|