%I #20 Nov 22 2020 12:17:28
%S 1,1,1,1,1,1,1,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,
%T 0,3,1,2,0,1,0,0,1,2,3,1,0,0,1,2,0,1,0,3,1,2,0,1,0,0,1,2,3,1,0,0,1,2,
%U 0,1,0,3,1,2,0,1,0,0,1,2,3,1,0,0,1,2,0,1,0,3,1,2,0,4,1,0,0,0,1,2,3,0,1,0,0,0
%N Irregular triangle read by rows: T(n,k) is the number of parts in the partition of n into k consecutive parts that differ by 6, n >= 1, k >= 1, and the first element of column k is in the row that is the k-th octagonal number (A000567).
%C Since the trivial partition n is counted, so T(n,1) = 1.
%C This is an irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in the row that is the k-th octagonal number.
%C This triangle can be represented with a diagram of overlapping curves, in which every column of triangle is represented by a periodic curve.
%C For a general theorem about the triangles of this family see A285914.
%F T(n,k) = k*A334946(n,k).
%e Triangle begins (rows 1..24).
%e 1;
%e 1;
%e 1;
%e 1;
%e 1;
%e 1;
%e 1;
%e 1, 2;
%e 1, 0;
%e 1, 2;
%e 1, 0;
%e 1, 2;
%e 1, 0;
%e 1, 2;
%e 1, 0;
%e 1, 2;
%e 1, 0;
%e 1, 2;
%e 1, 0;
%e 1, 2;
%e 1, 0, 3;
%e 1, 2, 0;
%e 1, 0, 0;
%e 1, 2, 3;
%e ...
%e For n = 24 there are three partitions of 24 into consecutive parts that differ by 6, including 24 as a valid partition. They are [24], [15, 9] and [14, 8, 2]. There are 1, 2 and 3 parts respectively, so the 24th row of this triangle is [1, 2, 3].
%Y Row sums give A334949.
%Y Triangles of the same family where the parts differ by d are A127093 (d=0), A285914 (d=1), A330466 (d=2), A330888 (d=3), A334462 (d=4), A334540 (d=5), this sequence (d=6).
%Y Cf. A000567, A334946, A334948, A334953.
%K nonn,tabf
%O 1,9
%A _Omar E. Pol_, May 27 2020