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%I #10 Aug 18 2017 03:14:34
%S 1,3,1,4,3,1,6,4,3,1,7,6,4,3,1,9,7,6,4,3,1,10,9,7,6,4,3,1,12,10,9,7,6,
%T 4,3,1,13,12,10,9,7,6,4,3,1,15,13,12,10,9,7,6,4,3,1,16,15,13,12,10,9,
%U 7,6,4,3,1,18,16,15,13,12,10,9,7,6,4,3,1,19,18,16,15,13,12,10,9,7,6,4,3,1
%N Triangle of numbers that are 0 or 1 mod 3.
%C The matrix operations (J * R), (R * J) are commutative since J * R = R * J.
%C Row sums = A006578.
%C Rows and columns of the triangle are all 0 or 1 mod 3 terms: A032766.
%C A104567 row sums also = A006578.
%C A006578(2n-1) = A001082(2n).
%F All columns (with offset); and all rows (starting from the right) are 0 or 1 mod 3 (A032766). Extract the triangle from the product J * R; J = [1; 2, 1; 1, 2, 1; 2, 1, 2, 1; ...]; R = [1; 1, 1; 1, 1, 1; ...] (infinite lower triangular matrices, with the rest zeros).
%e The first few rows are:
%e 1;
%e 3, 1;
%e 4, 3, 1;
%e 6, 4, 3, 1;
%e 7, 6, 4, 3, 1;
%e 9, 7, 6, 4, 3, 1;
%e ...
%p it:=array(1..1000): i:=1: for n from 1 to 1000 do if n mod 3 <> 2 then it[i]:=n; i:=i+1 fi: od: for j from 1 to 25 do for k from j to 1 by -1 do printf(`%d,`,it[k]) od: od: # _James A. Sellers_, Apr 09 2005
%Y Cf. A001082, A006578, A104566, A104567.
%K nonn,tabl
%O 0,2
%A _Gary W. Adamson_, Mar 16 2005
%E More terms from _James A. Sellers_, Apr 09 2005
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