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%I #24 Aug 03 2019 17:58:51
%S 1,2,1,5,3,1,15,9,4,1,52,31,14,5,1,203,121,54,20,6,1,877,523,233,85,
%T 27,7,1,4140,2469,1101,400,125,35,8,1,21147,12611,5625,2046,635,175,
%U 44,9,1,115975,69161,30846,11226,3488,952,236,54,10,1
%N Triangle read by rows, (1 / ((-1)*A129184 * A007318 + I)) - I, I = Identity matrix.
%C Inverse binomial transform of the triangle shifts to left (= adding I as right border, I = Identity matrix); resulting in reversed rows of A121207.
%C Left border = Bell numbers, A000110 = eigensequence of Pascal's triangle.
%C Successive columns from left to right = eigensequences of Pascal's triangle deleting columns one at a time.
%C Row sums of the triangle = A060719: (1, 3, 9, 29, 103, ...). - _Gary W. Adamson_, May 20 2013
%C From _Gary W. Adamson_, Jul 18 2019: (Start)
%C Rows are eigensequences of triangles exemplified by the following arrangement of binomial sequences. Example: row 5 is (1, 5, 14, 31, 52, 0, 0, 0, ...), the eigensequence of:
%C 1;
%C 4, 1;
%C 6, 3, 1;
%C 4, 3, 2, 1;
%C 1, 1, 1, 1, 1;
%C ... and the rest zeros.
%C Similarly, the production matrix for (1, 6, 20, 54, 121, 203, 0, 0, 0, ...) is:
%C 1;
%C 5, 1;
%C 10, 4, 1;
%C 10, 6, 3, 1;
%C 5, 4, 3, 2, 1;
%C 1, 1, 1, 1, 1, 1;
%C ... and the rest zeros. (End)
%F Triangle read by rows, 1 / ((-1)*A129184 * A051731 + I), I = Identity matrix.
%F Equals reversal by rows of triangle A121207, then delete right border. A121207 begins: 1; 1, 1; 1, 1, 2 1, 1, 3, 5; ...
%e First few rows of the triangle:
%e 1;
%e 2, 1;
%e 5, 3, 1;
%e 15, 9, 4, 1;
%e 52, 31, 14, 5, 1;
%e 203, 121, 54, 20, 6, 1;
%e 877, 523, 233, 85, 27, 7, 1;
%e 4140, 2469, 1101, 400, 125, 35, 8, 1;
%e 21147, 12611, 5625, 2046, 635, 175, 44, 9, 1;
%e 115975, 69161, 30846, 11226, 3488, 952, 236, 54, 10, 1;
%e ...
%Y Cf. A121207, A124496, A186020.
%Y Cf. A060719.
%K nonn,tabl
%O 0,2
%A _Gary W. Adamson_, May 03 2009
%E Corrected by _Alois P. Heinz_, Apr 18 2013
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