A180013 Triangular array read by rows: T(n,k) = number of fixed points in the permutations of {1,2,...,n} that have exactly k cycles; n>=1, 1<=k<=n.
1, 0, 2, 0, 3, 3, 0, 8, 12, 4, 0, 30, 55, 30, 5, 0, 144, 300, 210, 60, 6, 0, 840, 1918, 1575, 595, 105, 7, 0, 5760, 14112, 12992, 5880, 1400, 168, 8, 0, 45360, 117612, 118188, 60921, 17640, 2898, 252, 9, 0, 403200, 1095840, 1181240, 672840, 224490, 45360, 5460, 360, 10
Offset: 1
Examples
T(4,3)= 12 because there are 12 fixed points in the permutations of 4 that have 3 cycles: (1)(2)(4,3); (1)(3,2)(4); (1)(4,2)(3); (2,1)(3)(4); (3,1)(2)(4); (4,1)(2)(3) where the permutations are represented in their cycle notation. 1 0 2 0 3 3 0 8 12 4 0 30 55 30 5 0 144 300 210 60 6 0 840 1918 1575 595 105 7
Links
- G. C. Greubel, Table of n, a(n) for n = 1..5050
Programs
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Maple
egf:= k-> x * (log(1/(1-x)))^(k-1) / (k-1)!: T:= (n,k)-> n! * coeff(series(egf(k), x, n+1), x, n): seq(seq(T(n, k), k=1..n), n=1..10); # Alois P. Heinz, Jan 16 2011 # As coefficients of polynomials: with(PolynomialTools): with(ListTools): A180013_row := proc(n) `if`(n=0, 1,(n+1)!*hypergeom([-n,1-x],[1],1)); CoefficientList(simplify(%),x) end: FlattenOnce([seq(A180013_row(n), n=0..9)]); # Peter Luschny, Jan 28 2016
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Mathematica
Flatten[Table[Table[(n + 1) Abs[StirlingS1[n, k]], {k, 0, n}], {n, 0, 9}],1] (* Olivier Gérard, Oct 23 2012 *)
Formula
E.g.f.: for column k: x*(log(1/(1-x)))^(k-1)/(k-1)!.
T(n, k) = [x^k] (n+1)!*hypergeom([-n,1-x],[1],1) for n>0. - Peter Luschny, Jan 28 2016
Extensions
More terms from Alois P. Heinz, Jan 16 2011
Comments