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Triangle read by rows, A173108 * the diagonalized variant of A173110
3

%I #3 Mar 30 2012 17:25:35

%S 1,1,2,1,5,1,15,2,3,52,5,3,203,15,6,6,877,52,15,6,4140,203,45,12,20,

%T 21147,877,156,30,20,115975,4140,609,90,40,60,678570,21147,2631,312,

%U 100,60

%N Triangle read by rows, A173108 * the diagonalized variant of A173110

%C Row sums = A173110: (1, 1, 3, 6, 20, 60, 230, 950, 4420, 22230,...).

%F Let triangle A173108 = Q, and M = an infinite lower triangular matrix with A173110 as the rightmost diagonal and the rest zeros. Triangle A173111 = Q*M.

%e First few rows of the triangle =

%e 1;

%e 1;

%e 2, 1;

%e 5, 1;

%e 15, 2, 3;

%e 52, 5, 3;

%e 203, 15, 6, 6;

%e 877, 52, 15, 6;

%e 4140, 203, 45, 12, 20;

%e 21147, 877, 156, 30, 20;

%e 115975, 4140, 609, 90, 40, 60;

%e 678570, 21147, 2631, 312, 100, 60;

%e ...

%e Example: row 7 = termwise products of (877, 52, 5, 1) and (1, 1, 3, 6) =

%e (877, 52, 15, 6); where (877, 52, 5, 1) = row 7 of triangle A173108, and

%e (1, 1, 3, 6) = the first four terms of sequence A173109.

%Y A173108, A173109, A173110

%K nonn,tabl

%O 0,3

%A _Gary W. Adamson_, Feb 09 2010