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%I #6 Nov 20 2022 10:53:05
%S 1,1,1,2,3,2,6,11,12,7,24,50,70,70,36,120,274,450,595,540,250,720,
%T 1764,3248,5145,6300,5250,2229
%N Eigentriangle of A130534.
%C Right border = A143805 (1, 1, 2, 7, 36, 250,...) = row sums shifted one place to the left, = (1, 2, 7, 36, 250,...). Sum of n-th row terms = rightmost term of next row.
%C A130534 = the Stirling cycle numbers:
%C 1;
%C 1, 1;
%C 2, 3, 1;
%C 6, 11, 6, 1;
%C ...
%C The triangle by rows, applies termwise products of the eigensequence terms of A130534: (1, 1, 2, 7, 36, 250,...) = A143805; to row terms of A130534. Thus row 3 = (6, 11, 12, 7) = (6, 11, 6, 1) and termwise product of the first 4 terms of A143805: (1, 1, 2, 7).
%F Triangle read by rows, A130534 * (A143805 * 0^(n-k)); 0<=k<=n.
%e First few rows of the triangle:
%e 1;
%e 1, 1;
%e 2, 3, 2;
%e 6, 11, 12, 7;
%e 24, 50, 70, 70, 36;
%e 120, 274, 450, 595, 540, 250;
%e 720, 1764, 3248, 5145, 6300, 5250, 2229;
%e ...
%Y Cf. A130534, A143805.
%K nonn,tabl,more
%O 0,4
%A _Gary W. Adamson_, Sep 01 2008
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