|
Search: id:A000240
|
|
|
| A000240 |
|
Rencontres numbers: permutations with exactly one fixed point.
(Formerly M2763 N1111)
|
|
+0
12
|
|
| 1, 0, 3, 8, 45, 264, 1855, 14832, 133497, 1334960, 14684571, 176214840, 2290792933, 32071101048, 481066515735, 7697064251744, 130850092279665, 2355301661033952, 44750731559645107, 895014631192902120, 18795307255050944541, 413496759611120779880
(list; graph; listen)
|
|
|
OFFSET
|
1,3
|
|
|
REFERENCES
|
S. K. Das and N. Deo, Rencontres graphs: a family of bipartite graphs, Fib. Quart., Vol. 25, No. 3, August 1987, 250-262.
I. Kaplansky, Symbolic solution of certain problems in permutations, Bull. Amer. Math. Soc., 50 (1944), 906-914.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 65.
|
|
LINKS
|
T. D. Noe, Table of n, a(n) for n=1..100
|
|
FORMULA
|
E.g.f. = exp(-x)*(1+x^3)/(1-x)(1-x^2). a(n)=sum((-1)^k*n!/k!, k=0..n-1).
a(n) = n*a(n-1)-(-1)^n*n = A000166(n)-(-1)^n = n*A000166(n-1) = A000387(n+1)*2/(n+1) = A000449(n+2)*6/((n+1)*(n+2))
|
|
EXAMPLE
|
a(3)=3 because the permutations of (1,2,3) with one fixed point are (1,3,2), (3,2,1) and (2,1,3)
|
|
MAPLE
|
a:=n->sum(n!*sum((-1)^k/k!, j=0..n), k=0..n): seq(a(n), n=0..21); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 17 2007
|
|
CROSSREFS
|
Cf. A008290, A000166, A000387, etc.
A diagonal of A008291.
Sequence in context: A074435 A039647 A071533 this_sequence A132103 A040018 A019016
Adjacent sequences: A000237 A000238 A000239 this_sequence A000241 A000242 A000243
|
|
KEYWORD
|
nonn,easy,nice
|
|
AUTHOR
|
njas, Simon Plouffe (plouffe(AT)math.uqam.ca)
|
|
|
Search completed in 0.002 seconds
|
