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%I #14 Jun 04 2022 22:03:59
%S 1,1,2,3,1,5,1,7,2,11,3,1,15,5,1,22,7,2,30,11,3,1,42,15,5,1,56,22,7,2,
%T 77,30,11,3,1,101,42,15,5,1,135,56,22,7,2,176,77,30,11,3,1,231,101,42,
%U 15,5,1,297,135,56,22,7,2,385,176,77,30,11,3,1
%N Triangle by columns, A000041 shifted down thrice, k>=0.
%C Row sums = A024787, the numbers of 3's in all partitions of n, where A024787 starts with offset 1: (0, 0, 1, 1, 2, 4, 6, 9, 15,...). Triangle A173239 row sums start with the first "1" of A024787.
%C Let the triangle = M as an infinite lower triangular matrix. Then Lim_{n->inf} = A173241, the Euler transform of A051064 (the ruler function for 3).
%C Let P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), then P(x) = A(x) / A(x^3), where A(x) = polcoeff A173241: (1 + x + 2x^2 + 4x^3 + 6x^4 + ...)
%C Refer to A173238 comments for three conjectures relating A000041 to the infinite set of generalized ruler function sequences.
%F T(n,k) = A000041(n-3*k) for k=0..floor(n/3).
%e First few rows of the triangle =
%e 1;
%e 1;
%e 2;
%e 3, 1;
%e 5, 1;
%e 7, 2;
%e 11, 3, 1;
%e 15, 5, 1;
%e 22, 7, 2;
%e 30, 11, 3, 1;
%e 42, 15, 5, 1;
%e 56, 22, 7, 2;
%e 77, 30, 11, 3, 1;
%e 101, 42, 15, 5, 1;
%e 135, 56, 22, 7, 2;
%e 176, 77, 30, 11, 3, 1;
%e 231, 101, 42, 15, 5, 1;
%e 297, 135, 56, 22, 7, 2;
%e 385, 176, 77, 30, 11, 3, 1;
%e 490, 231, 101, 42, 15, 5, 1;
%e 627, 297, 135, 56, 22, 7, 2;
%e 792, 385, 176, 77, 30, 11, 3, 1;
%e 1002,490, 231, 101, 42, 15, 5, 1;
%e 1255, 627, 297, 135, 56, 22, 7, 2;
%e 1575, 792, 385, 176, 77, 30, 11, 3, 1;
%e ...
%Y Cf. A000041, A173238, A173241, A051064, A024787.
%K nonn,tabf,easy
%O 0,3
%A _Gary W. Adamson_, Feb 13 2010
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