%I
%S 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,
%T 595,105,7,0,5760,14112,12992,5880,1400,168,8,0,45360,117612,118188,
%U 60921,17640,2898,252,9,0,403200,1095840,1181240,672840,224490,45360,5460,360,10
%N 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.
%C Row sums = n! which is the number of fixed points in all the permutations of {1,2,...,n}.
%C It appears that column k = 2 is A001048 (with different offset).
%C From _Olivier Gérard_, Oct 23 2012: (Start)
%C This is a multiple of the triangle of Stirling numbers of the first kind, A180013(n,k) = (n)*A132393(n1,k).
%C 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).
%C (End)
%H G. C. Greubel, <a href="/A180013/b180013.txt">Table of n, a(n) for n = 1..5050</a>
%F E.g.f.: for column k: x*(log(1/(1x)))^(k1)/(k1)!.
%F T(n, k) = [x^k] (n+1)!*hypergeom([n,1x],[1],1)) for n>0. # _Peter Luschny_, Jan 28 2016
%e 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.
%e 1
%e 0 2
%e 0 3 3
%e 0 8 12 4
%e 0 30 55 30 5
%e 0 144 300 210 60 6
%e 0 840 1918 1575 595 105 7
%p egf:= k> x * (log(1/(1x)))^(k1) / (k1)!:
%p T:= (n,k)> n! * coeff(series(egf(k), x, n+1), x, n):
%p seq(seq(T(n, k), k=1..n), n=1..10); # _Alois P. Heinz_, Jan 16 2011
%p # As coefficients of polynomials:
%p with(PolynomialTools): with(ListTools): A180013_row := proc(n)
%p `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
%t Flatten[Table[Table[(n + 1) Abs[StirlingS1[n, k]], {k, 0, n}], {n, 0, 9}],1] (* _Olivier Gérard_, Oct 23 2012 *)
%Y Cf. A000142, A001048. Diagonal, lower diagonal give: A000027, A027480(n+1).
%K nonn,tabl
%O 1,3
%A _Geoffrey Critzer_, Jan 13 2011
%E More terms from _Alois P. Heinz_, Jan 16 2011
