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%I #24 Jul 23 2024 03:47:07
%S 1,1,1,1,2,1,1,3,6,1,1,4,15,24,1,1,5,28,105,116,1,1,6,45,280,929,648,
%T 1,1,7,66,585,3600,9851,4088,1,1,8,91,1056,9865,56240,121071,28640,1
%N Array T(n,k) read by antidiagonals: the k-th column contains the first column of the k-th power of A039755.
%F Let A039755 (an analog of Stirling numbers of the second kind) be an infinite lower triangular matrix M; then the vector M^k * [1, 0, 0, 0, ...] (first column of the k-th power) is the k-th column of this array.
%e 1 1 1 1 1 1 1 1
%e 1 2 3 4 5 6 7 8
%e 1 6 15 28 45 66 91 120
%e 1 24 105 280 585 1056 1729 2640
%e 1 116 929 3600 9865 22036 43001 76224
%e 1 648 9851 56240 203565 565096 1318023 2717856
%e 1 4088 121071 1029920 4953205 17148936 47920803 115146816
%e 1 28640 1685585 21569600 138529105 600001696 2012844225 5644055040
%p A111670 := proc(n,k)
%p local A,i,j ;
%p A := Matrix(n,n) ;
%p for i from 1 to n do
%p for j from 1 to n do
%p A[i,j] := A039755(i-1,j-1) ;
%p end do:
%p end do:
%p LinearAlgebra[MatrixPower](A,k) ;
%p %[n,1] ;
%p end proc:
%p for d from 2 to 12 do
%p for n from 1 to d-1 do
%p printf("%d,",A111670(n,d-n)) ;
%p end do:
%p end do: # _R. J. Mathar_, Jan 27 2023
%t nmax = 10;
%t A[n_, k_] := Sum[(-1)^(k-j)*(2j+1)^n*Binomial[k, j], {j, 0, k}]/(2^k*k!);
%t A039755 = Array[A, {nmax, nmax}, {0, 0}];
%t T = Table[MatrixPower[A039755, n][[All, 1]], {n, 1, nmax}] // Transpose;
%t Table[T[[n-k+1, k]], {n, 1, nmax}, {k, n, 1, -1}] // Flatten (* _Jean-François Alcover_, Apr 02 2024 *)
%Y Cf. A039755, A007405 (column 2), A000384 (row 2), A011199 (row 3).
%K nonn,tabl
%O 1,5
%A _Gary W. Adamson_, Aug 14 2005
%E Definition simplified by _R. J. Mathar_, Jan 27 2023