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A130478 Triangle T(n,k) = n! / A130477(n,k).

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%I A130478 #13 Jan 27 2025 07:20:12
%S A130478 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 A130478 3,2,40320,5760,840,144,30,8,3,2,362880,45360,5760,840,144,30,8,3,2,
%U A130478 3628800,403200,45360,5760,840,144,30,8,3,2
%N A130478 Triangle T(n,k) = n! / A130477(n,k).
%C A130478 Sums of reciprocals of rows is 1. - _Henry Bottomley_, Nov 05 2009
%F A130478 T(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.
%F A130478 Triangle by rows = n! followed by the first (n-1) reversed terms of A001048: (2, 3, 8, 30, 144, 840, ...).
%F A130478 Left border = (1, 2, 6, 24, 120, ...); while all other columns = A001048: (2, 3, 8, 30, ...).
%F A130478 n-th row of the triangle = n terms of: (n!; (n-1)!+(n-2)!; (n-2)!+(n-3)!; ...; 1! + 0!).
%e A130478 First few rows of the triangle:
%e A130478      1;
%e A130478      2,   2;
%e A130478      6,   3,   2;
%e A130478     24,   8,   3,  2;
%e A130478    120,  30,   8,  3, 2;
%e A130478    720, 144,  30,  8, 3, 2;
%e A130478   5040, 840, 144, 30, 8, 3, 2;
%e A130478   ...
%e A130478 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 A130478 Row 5 = (120, 30, 8, 3, 2) = 5! + (4!+3!) + (3!+2!) + (2!+1!) + (1!+1).
%e A130478 Row 5 = 120 followed by the first reversed 4 terms of A001048; i.e., 120 followed by 30, 8, 3, 2.
%Y A130478 Cf. A130493 (row sums), A001048, A130493, A130477.
%K A130478 nonn,tabl
%O A130478 1,2
%A A130478 _Gary W. Adamson_, May 31 2007
%E A130478 Corrected and extended by _Henry Bottomley_, Nov 05 2009