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%I #32 May 04 2023 10:05:33
%S 1,2,1,6,2,1,24,4,2,1,120,12,4,2,1,720,36,8,4,2,1,5040,144,24,8,4,2,1,
%T 40320,576,72,16,8,4,2,1,362880,2880,216,48,16,8,4,2,1,3628800,14400,
%U 864,144,32,16,8,4,2,1
%N Triangle read by rows: T(n,k) is the number of permutations of length n with each cycle of the permutation containing only elements that are identical (mod k), where 1 <= k <= n.
%C Let [n] denote {1,2,...,n} and let [n](j,k) denote the subset of [n] consisting of all elements of [n] that equal j mod k. The cardinality of [n](j,k) equals ceiling(n/k) for j = 1..(n mod k) and equals floor(n/k) for j > (n mod k). Therefore, upon permuting the elements of each [n](j,k) subset, we obtain T(n,k) = ceiling(n/k)!)^(n mod k)*(floor(n/k)!)^(k-n mod k).
%H Per Alexandersson, Frether Getachew Kebede, Samuel Asefa Fufa, and Dun Qiu, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Getachew/get3.html">Pattern-Avoidance and Fuss-Catalan Numbers</a>, J. Int. Seq. (2023) Vol. 26, Art. 23.4.2.
%F T(n,k) = (ceiling(n/k)!)^(n mod k)*(floor(n/k)!)^(k-n mod k) for 1 <= k <= n.
%F T(n,1) = A000142(n).
%F T(n,2) = A010551(n) for n > 1.
%F T(n,3) = A264557(n) for n > 2.
%F T(n,4) = A264635(n) for n > 3.
%F T(n,5) = A264656(n) for n > 4.
%F T(n,k) = Product_{i=0..k-1} floor((n+i)/k)!. - _Alois P. Heinz_, May 23 2020
%e Triangle begins:
%e 1;
%e 2 1;
%e 6 2 1;
%e 24 4 2 1;
%e 120 12 4 2 1;
%e ...
%e T(6,3) counts the 8 permutations of [6] where all cycle-mates are identical mod 3, namely, (1 4)(2 5)(3 6), (1 4)(2 5)(3)(6), (1 4)(2)(5)(3 6), (1)(4)(2 5)(3 6), (1 4)(2)(5)(3)(6), (1)(4)(2 5)(3)(6), (1)(4)(2)(5)(3 6) and (1)(2)(3)(4)(5)(6).
%p seq(seq((ceil(n/k)!)^(n mod k)*(floor(n/k)!)^(k-(n mod k)), k=1..n), n=1..10);
%t Table[(Ceiling[n/k]!)^Mod[n, k]*(Floor[n/k]!)^(k - Mod[n, k]), {n, 10}, {k, n}] // Flatten (* _Michael De Vlieger_, Jun 28 2020 *)
%Y Cf. A000142, A010551, A264557, A264635, A264656.
%Y Cf. A275062.
%K nonn,tabl
%O 1,2
%A _Dennis P. Walsh_, May 23 2020
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