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A137775
Number of triples of permutations on n letters such that for each j, exactly one of the permutations fixes j and the other two have the same image on j.
9
1, 0, 3, 6, 45, 252, 1935, 16146, 153657, 1616760, 18699579, 235498590, 3207570597, 46968796404, 735689606535, 12272343940458, 217191191400945, 4064131571557104, 80166987477918963, 1662468879466624950, 36156426996107254941, 822876672690142595820
OFFSET
0,3
COMMENTS
This sequence arises in a calculation of the fourth moments of the volumes of random polytopes in certain very symmetric convex bodies.
REFERENCES
M. Meckes, Volumens of symmetric random polytopes, Arch. Math. 82 (2004) 85--96.
LINKS
FORMULA
a(n) = (n-1) * (a(n-1) + 3*a(n-2)) with a(0)=1. [corrected by Seiichi Manyama, Apr 23 2025]
E.g.f.: exp(-3x)/(1-x)^3.
a(n) is the number of derangements (permutations with no fixed points) of n elements where each cycle is colored with one of three colors. - Michael Somos, Jan 19 2011
G.f.: hypergeom([1,3],[],x/(1+3*x))/(1+3*x). - Mark van Hoeij, Nov 08 2011
a(n) ~ n! * exp(-3) * n^2/2. - Vaclav Kotesovec, Oct 08 2013
a(n) = n! * Sum_{k=0..n} (-3)^(n-k) * binomial(k+2,2)/(n-k)!. - Seiichi Manyama, Apr 23 2025
EXAMPLE
a(2) = 3 because one of the permutations must be the identity and the other two are the transposition (1 2); there are three ways to pick which is the identity.
a(4) = 45 because there are 6 derangements with one 4-cycle with 3^1 ways to color each derangement and 3 derangements with two 2-cycles with 3^2 ways to color each derangement. - Michael Somos, Jan 19 2011
MATHEMATICA
Range[0, 20]! CoefficientList[Series[Exp[ -3x]/(1 - x)^3, {x, 0, 20}], x]
PROG
(PARI) {a(n) = if( n<0, 0, n! * polcoeff( exp( -3 * x + x * O(x^n) ) / ( 1 - x )^3, n ) )} /* Michael Somos, Jan 19 2011 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Mark W. Meckes (mark.meckes(AT)case.edu), May 06 2008
EXTENSIONS
Added a(0)=1 by Michael Somos, Jan 19 2011
STATUS
approved