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A368213 Triangular array read by rows: Number of permutations of [n] that factor into exactly k-cycles, ordered by n (rows) and divisors k of n (columns).

%I A368213 #49 Dec 18 2023 16:33:12
%S A368213 1,1,1,1,0,2,1,3,0,6,1,0,0,0,24,1,15,40,0,0,120,1,0,0,0,0,0,720,1,105,
%T A368213 0,1260,0,0,0,5040,1,0,2240,0,0,0,0,0,40320,1,945,0,0,72576,0,0,0,0,
%U A368213 362880,1,0,0,0,0,0,0,0,0,0,3628800,1,10395,246400,1247400,0,6652800,0,0,0,0,0,39916800
%N A368213 Triangular array read by rows: Number of permutations of [n] that factor into exactly k-cycles, ordered by n (rows) and divisors k of n (columns).
%D A368213 P. Flajolet and R. Sedgewick, Analytic Combinatorics, Cambridge University Press, 2009, pages 120-122.
%H A368213 Marko Riedel, math.stackexchange.com, <a href="https://math.stackexchange.com/questions/4828858/">Number of permutations in the symmetric group</a>.
%H A368213 Marko Riedel, <a href="/A368213/a368213_1.pdf">Computing the number of permutations of [n] with k m-cycles with analytic combinatorics.</a>
%H A368213 Marko Riedel, <a href="/A368213/a368213_2.pdf">Computing the number of permutations of [n] with k m-cycles using the principle of inclusion-exclusion.</a>
%F A368213 T(n, k) = n! / ( k^(n/k) * (n/k)! ) if k divides n otherwise 0.
%e A368213 Row n=6 is 1, 15, 40, 120 because there is one permutation of [6] consisting of six fixed points, there are 15 permutations consisting of three transpositions, there are forty permutations consisting of two three-cycles and there are one hundred and twenty permutations consisting of just one six-cycle (6!/6).
%e A368213 Triangular array starts:
%e A368213 [ 1] 1;
%e A368213 [ 2] 1,   1;
%e A368213 [ 3] 1,   0,    2;
%e A368213 [ 4] 1,   3,    0,    6;
%e A368213 [ 5] 1,   0,    0,    0,    24;
%e A368213 [ 6] 1,  15,   40,    0,     0, 120;
%e A368213 [ 7] 1,   0,    0,    0,     0,   0, 720;
%e A368213 [ 8] 1, 105,    0, 1260,     0,   0,   0, 5040;
%e A368213 [ 9] 1,   0, 2240,    0,     0,   0,   0,    0, 40320;
%e A368213 [10] 1, 945,    0,    0, 72576,   0,   0,    0,     0, 362880;
%p A368213 T:= (n, m)-> `if`(irem(n,m)=0, n!/m^(n/m)/(n/m)!, 0):
%p A368213 seq(seq(T(n, m), m = 1..n), n=1..15);
%t A368213 A368213[n_,k_]:=If[Divisible[n,k],n!/(k^(n/k)(n/k)!),0];
%t A368213 Table[A368213[n,k],{n,15},{k,n}] (* _Paolo Xausa_, Dec 18 2023 *)
%o A368213 (SageMath)
%o A368213 def T(n, d): return factorial(n) // (d ** (n//d) * factorial(n//d))
%o A368213 for n in range(1, 19):
%o A368213     print([T(n, d) if n % d == 0 else 0 for d in range(1, n+1)])
%o A368213 # _Peter Luschny_, Dec 17 2023
%Y A368213 Cf. A005225 (row sums), A008290.
%Y A368213 Cf. A123023 (column 2), A052502 (column 3), A060706 (column 4).
%K A368213 nonn,tabl
%O A368213 1,6
%A A368213 _Marko Riedel_, Dec 17 2023