

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



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
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OFFSET

1,3


COMMENTS

Row sums = n! which is the number of fixed points in all the permutations of {1,2,...,n}.
It appears that column k = 2 is A001048 (with different offset).
From Olivier Gérard, Oct 23 2012: (Start)
This is a multiple of the triangle of Stirling numbers of the first kind, A180013(n,k) = (n)*A132393(n1,k).
Another interpretation is : T(n,nk) is the total number of ways to insert the symbol n among the cycles of permutations of [n1] with (n+1k) cycles to form a canonical cycle representation of a permutation of [n]. For each cycle of length c, there are c places to insert a symbol, and for each permutation there is the possibility to create a new cycle (a fixed point).
(End)


LINKS

G. C. Greubel, Table of n, a(n) for n = 1..5050


FORMULA

E.g.f.: for column k: x*(log(1/(1x)))^(k1)/(k1)!.
T(n, k) = [x^k] (n+1)!*hypergeom([n,1x],[1],1)) for n>0. # Peter Luschny, Jan 28 2016


EXAMPLE

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


MAPLE

egf:= k> x * (log(1/(1x)))^(k1) / (k1)!:
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, 1x], [1], 1)); CoefficientList(simplify(%), x) end: FlattenOnce([seq(A180013_row(n), n=0..9)]); # Peter Luschny, Jan 28 2016


MATHEMATICA

Flatten[Table[Table[(n + 1) Abs[StirlingS1[n, k]], {k, 0, n}], {n, 0, 9}], 1] (* Olivier Gérard, Oct 23 2012 *)


CROSSREFS

Cf. A000142, A001048. Diagonal, lower diagonal give: A000027, A027480(n+1).
Sequence in context: A134409 A327878 A298605 * A094067 A094112 A326926
Adjacent sequences: A180010 A180011 A180012 * A180014 A180015 A180016


KEYWORD

nonn,tabl


AUTHOR

Geoffrey Critzer, Jan 13 2011


EXTENSIONS

More terms from Alois P. Heinz, Jan 16 2011


STATUS

approved



