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Triangle T(n,m) (1<=m<=n) giving number of maps f:N -> N such that f^m(X)=X+n for all natural numbers X.
2

%I #31 Feb 05 2019 17:17:46

%S 1,1,2,1,0,6,1,12,0,24,1,0,0,0,120,1,120,360,0,0,720,1,0,0,0,0,0,5040,

%T 1,1680,0,20160,0,0,0,40320,1,0,60480,0,0,0,0,0,362880,1,30240,0,0,

%U 1814400,0,0,0,0,3628800,1,0,0,0,0,0,0,0,0,0,39916800

%N Triangle T(n,m) (1<=m<=n) giving number of maps f:N -> N such that f^m(X)=X+n for all natural numbers X.

%H Vincenzo Librandi, <a href="/A066387/b066387.txt">Rows n = 1..100, flattened</a>

%H A. Heinis, R. Jeurissen and L. Kamstra, <a href="http://www.nieuwarchief.nl/serie5/pdf/naw5-2001-02-2-189.pdf">Problem 18</a> and <a href="http://www.nieuwarchief.nl/serie5/pdf/naw5-2001-02-4-380.pdf">solution</a>, Nieuw Arch. Wisk. 5/2 (2001) 380.

%F T(n,m) = n!/(n/m)! if m|n, T(n,m) = 0 otherwise.

%e Triangle T(n,m) begins:

%e 1;

%e 1, 2;

%e 1, 0, 6;

%e 1, 12, 0, 24;

%e 1, 0, 0, 0, 120;

%e 1, 120, 360, 0, 0, 720;

%e 1, 0, 0, 0, 0, 0, 5040;

%e 1, 1680, 0, 20160, 0, 0, 0, 40320;

%e ...

%t t[n_, m_] /; Divisible[n, m] := n!/(n/m)!; t[_, _] = 0; Flatten[Table[t[n, m], {n, 1, 11}, {m, 1, n}]] (* _Jean-François Alcover_, Nov 29 2011 *)

%Y Row sums give A057625.

%Y Main diagonal gives A000142.

%Y m-section of column m=2-4 (for n>0) gives: A001813, A064350, A166338.

%K easy,nonn,tabl,nice

%O 1,3

%A _Floor van Lamoen_, Dec 23 2001