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%I #6 Feb 21 2022 00:16:00
%S 1,2,2,6,3,2,24,8,3,2,120,30,8,3,2,720,144,30,8,3,2,5040,840,144,30,8,
%T 3,2,40320,5760,840,144,30,8,3,2,362880,45360,5760,840,144,30,8,3,2,
%U 3628800,403200,45360,5760,840,144,30,8,3,2
%N Triangle T(n,k) = n! / A130477(n,k).
%C Row sums = A130494: (1, 4, 11, 37, 163, ...).
%C Sums of reciprocals of rows is 1 - _Henry Bottomley_, Nov 05 2009
%F Triangle(n,k) = n! / A130477(n,k); such that by rows as vector terms, (n-th row of A130477) dot (n-th row of A130478) = n-th row of A130493 = n! repeated n times. Triangle A130478 by rows = n! followed by the first (n-1)reversed terms of A001048: (2, 3, 8, 30, 144, 840, ...). Left border = (1, 2, 6, 24, 120, ...); while all other columns = A001048: (2, 3, 8, 30, ...). n-th row of the triangle = n terms of: (n!; (n-1!)+(n-2!); (n-2!)+(n-3!); ... + (1! + 1).
%e First few rows of the triangle:
%e 1;
%e 2, 2;
%e 6, 3, 2;
%e 24, 8, 3, 2;
%e 120, 30, 8, 3, 2;
%e 720, 144, 30, 8, 3, 2;
%e 5040, 840, 144, 30, 8, 3, 2;
%e ...
%e Row 4 = (24, 8, 3, 2), terms such that (24, 8, 3, 2) dot (1, 3, 8, 12) = (24, 24, 24, 24), where (1, 3, 8, 12) = row 4 of A130477 and (24, 24, 24, 24) = row 4 of A130493.
%e Row 5 = (120, 30, 8, 3, 2) = 5! + (4!+3!) + (3!+2!) + (2!+1!) + (1!+1).
%e Row 5 = 120 followed by the first reversed 4 terms of A001048; i.e., 120 followed by 30, 8, 3, 2.
%Y Cf. A130493, A001048, A130493, A130477.
%K nonn,tabl
%O 1,2
%A _Gary W. Adamson_, May 31 2007
%E Corrected and extended by _Henry Bottomley_, Nov 05 2009
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