%I #10 Jun 13 2017 21:49:29
%S 1,2,1,6,3,1,24,11,4,1,120,49,18,5,1,720,261,92,27,6,1,5040,1631,536,
%T 159,38,7,1,40320,11743,3552,1029,256,51,8,1,362880,95901,26608,7353,
%U 1848,389,66,9,1,3628800,876809,223456,58095,14384,3125,564,83,10,1
%N Square array, read by antidiagonals, where the n-th row is the n-th binomial transform of the factorials, starting with row 0: {1!,2!,3!,...}.
%C Row 1 is A001339, antidiagonal sums form A089902 and the main diagonal is A089901; the next lower diagonal forms {1,4,27,256,..,n^n,..}, which is the hyperbinomial transform (cf. A088956) of the main diagonal.
%F T(0, k)=(k+1)!, T(n+1, n)=(n+1)^(n+1), T(n, k)=sum_{i=0..k}n^(k-i)*binomial(k, i)*(i+1)!
%F E.g.f.: 1/((1-y*exp(x))*(1-x)^2). E.g.f. (n-th row): exp(n*x)/(1-x)^2.
%e Note secondary diagonal: {(n+1)^(n+1)}; rows begin:
%e 1, 2,. 6,. 24,. 120,.. 720,.. 5040,..
%e 1, 3, 11,. 49,. 261,. 1631,. 11743,..
%e 1,_4, 18,. 92,. 536,. 3552,. 26608,..
%e 1, 5,_27, 159, 1029,. 7353,. 58095,..
%e 1, 6, 38,_256, 1848, 14384, 121264,..
%e 1, 7, 51, 389,_3125, 26595, 241015,..
%e 1, 8, 66, 564, 5016,_46656, 456048,..
%e 1, 9, 83, 787, 7701, 78077,_823543,..
%t t[n_, k_] := (n^(k+2) - Exp[n]*(n-k-1)*Gamma[k+2, n])/(k+1) // Round; Table[t[n-k, k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* _Jean-François Alcover_, Jun 24 2013 *)
%o (PARI) T(n,k)=if(n<0 || k<0,0,sum(i=0,k,n^(k-i)*binomial(k,i)*(i+1)!))
%Y Cf. A001339, A088956, A089901, A089902.
%K nonn,tabl
%O 0,2
%A _Paul D. Hanna_, Nov 14 2003