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%I #8 Feb 08 2022 22:44:16
%S 1,1,1,2,0,1,3,1,0,1,4,1,1,0,1,5,2,2,1,0,1,7,2,3,1,1,0,1,9,4,3,3,1,1,
%T 0,1,12,5,5,3,3,0,1,0,1,15,8,6,5,3,2,1,1,0,1,19,10,9,6,5,2,2,1,1,0,1,
%U 25,13,12,10,5,5,2,2,1,1,0,1,31,17,16,12,9,5,4,2,2,1,1,0,1
%N Triangle, row sums = A000041 starting (1, 2, 3, 5, 7, ...); derived from finite differences of p(x) = A(x)*A(x^2) = B(x)*B(x^3) = C(x)*C(x^4) = ...
%C Row sums = A000041 starting with offset 1: (1, 2, 3, 5, 7, 11, ...).
%F Given an array of rows satisfying p(x) = A(x)*A(x^2) = row 1 = A174065; row = 2 A174068 satisfying p(x) = B(x)*B(x^3); row 3 satisfies p(x) = C(x)*C(x^4), ... and so on; take finite differences from the top, becoming rows of triangle A174067.
%e First few rows of the array:
%e 1, 1, 1, 2, 3, 4, 5, 7, 9, 12, 15, 19, ... = A174065
%e 1, 1, 2, 2, 4, 5, 7, 9, 13, 17, 23, 29, ... = A174068
%e 1, 1, 2, 3, 4, 6, 9, 12, 16, 22, 29, 38, ... satisfies p(x) = C(x)*C(x^4)
%e 1, 1, 2, 3, 5, 6, 10, 13, 19, 25, 34, 44, ... analogous for k=5
%e 1, 1, 2, 3, 5, 7, 10, 14, 20, 28, 37, 49, ..................k=6
%e 1, 1, 2, 3, 5, 7, 11, 14, 21, 28, 39, 51, ..................k=7
%e 1, 1, 2, 3, 5, 7, 11, 15, 21, 29, 40, 53, ..................k=8
%e 1, 1, 2, 3, 5, 7, 11, 15, 22, 29, 41, 54, ..................k=9
%e 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 41, 55, ..................k=10
%e ...
%e Finally, take finite differences from the top, deleting the first 1, to obtain triangle A174067:
%e 1;
%e 1, 1;
%e 2, 0, 1;
%e 3, 1, 0, 1;
%e 4, 1, 1, 0, 1;
%e 5, 2, 2, 1, 0, 1;
%e 7, 2, 3, 1, 1, 0, 1;
%e 9, 4, 3, 3, 1, 1, 0, 1;
%e 12, 5, 5, 3, 3, 0, 1, 0, 1;
%e 15, 8, 6, 5, 3, 2, 1, 1, 0, 1;
%e 19, 10, 9, 6, 5, 2, 2, 1, 1, 0, 1;
%e 25, 13, 12, 10, 5, 5, 2, 2, 1, 1, 0, 1;
%e 31, 17, 16, 12, 9, 5, 4, 2, 2, 1, 1, 0, 1;
%e 38, 24, 20, 18, 11, 8, 5, 4, 2, 2, 1, 1, 0, 1;
%e ...
%Y Cf. A000041, A174065, A174066, A174068.
%K nonn,tabl
%O 1,4
%A _Gary W. Adamson_, Mar 06 2010