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A101817 Triangle read by rows: T(n,h) = number of functions f:{1,2,...,n}->{1,2,...,n} such that |Image(f)|=h; h=1,2,...,n, n=1,2,3,... . Essentially A090657, but without zeros.

%I A101817 #15 Mar 03 2022 12:43:49
%S A101817 1,2,2,3,18,6,4,84,144,24,5,300,1500,1200,120,6,930,10800,23400,10800,
%T A101817 720,7,2646,63210,294000,352800,105840,5040,8,7112,324576,2857680,
%U A101817 7056000,5362560,1128960,40320,9,18360,1524600,23496480,105099120
%N A101817 Triangle read by rows: T(n,h) = number of functions f:{1,2,...,n}->{1,2,...,n} such that |Image(f)|=h; h=1,2,...,n, n=1,2,3,... . Essentially A090657, but without zeros.
%C A101817 Row sums = n^n. T(n,1) = n, T(n,n) = n!.
%D A101817 H. Picquet, Note #124, L'Intermédiaire des Mathématiciens, 1 (1894), pp. 125-127. - _N. J. A. Sloane_, Feb 28 2022
%F A101817 T(n, h) = C(n, h)*U(n, h), where U(n, h) is the array in A019538. Thus T(n, h) = C(n, h)*h!*S(n, h), where S(n, h) is a Stirling number of the second kind (given by A048993 with zeros removed).
%F A101817 T(2n,n) = A288312(n). - _Alois P. Heinz_, Jun 07 2017
%e A101817 First rows:
%e A101817 1;
%e A101817 2,   2;
%e A101817 3,  18,   6;
%e A101817 4,  84, 144,  24;
%t A101817 Table[Table[StirlingS2[n, k] Binomial[n, k] k!, {k, 1, n}], {n, 1, 8}] // Grid
%Y A101817 Cf. A090657, A048993, A101818, A101819, A101821, A288312.
%K A101817 nonn,tabl
%O A101817 1,2
%A A101817 _Clark Kimberling_, Dec 17 2004