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Triangle T, read by rows, equal to a diagonal bisection of A118032 such that diagonal n of T equals diagonal 2n+1 of A118032: T(n,k) = A118032(2n+1-k,k); also equals the matrix product of A118032 and SHIFT_UP(A118032).
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%I #3 Mar 30 2012 18:36:56

%S 1,3,2,9,8,3,26,28,15,4,73,86,57,24,5,191,250,192,96,35,6,500,696,567,

%T 356,145,48,7,1234,1824,1683,1060,590,204,63,8,3051,4754,4392,3344,

%U 1765,906,273,80,9,7201,11562,12084,8672,5895,2718,1316,352,99,10,16995

%N Triangle T, read by rows, equal to a diagonal bisection of A118032 such that diagonal n of T equals diagonal 2n+1 of A118032: T(n,k) = A118032(2n+1-k,k); also equals the matrix product of A118032 and SHIFT_UP(A118032).

%e Triangle begins:

%e 1;

%e 3, 2;

%e 9, 8, 3;

%e 26, 28, 15, 4;

%e 73, 86, 57, 24, 5;

%e 191, 250, 192, 96, 35, 6;

%e 500, 696, 567, 356, 145, 48, 7;

%e 1234, 1824, 1683, 1060, 590, 204, 63, 8;

%e 3051, 4754, 4392, 3344, 1765, 906, 273, 80, 9;

%e 7201, 11562, 12084, 8672, 5895, 2718, 1316, 352, 99, 10; ...

%e which is formed from the odd-indexed diagonals of triangle

%e A118032, which starts:

%e 1;

%e 1, 1;

%e 2, 2, 1;

%e 3, 4, 3, 1;

%e 6, 8, 6, 4, 1;

%e 9, 14, 15, 8, 5, 1; ...

%e Let U = SHIFT_UP(A118032), shifting columns of A118032 up 1 row

%e and dropping the main diagonal, so that U =

%e 1;

%e 2, 2;

%e 3, 4, 3;

%e 6, 8, 6, 4;

%e 9, 14, 15, 8, 5;

%e 16, 28, 24, 24, 10, 6; ...

%e Then the matrix product A118032*U equals this triangle.

%Y . columns: A118046, A118047, A118048; A118049 (row sums); related triangles: A118032, A118040.

%K nonn,tabl

%O 0,2

%A _Paul D. Hanna_, Apr 10 2006