%I #19 Nov 22 2020 12:17:06

%S 1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,

%T 0,1,1,1,0,1,0,0,1,1,1,1,0,0,1,1,0,1,0,1,1,1,0,1,0,0,1,1,1,1,0,0,1,1,

%U 0,1,0,1,1,1,0,1,0,0,1,1,1,1,0,0,1,1,0,1,0,1,1,1,0,1,1,0,0,0,1,1,1,0,1,0,0

%N Irregular triangle read by rows: T(n,k) is the number of partitions of n into k consecutive parts that differ by 6, and the first element of column k is in the row that is the k-th octagonal number (A000567).

%C T(n,k) is 0 or 1, so T(n,k) represents the "existence" of the mentioned partition: 1 = exists, 0 = does not exist.

%C Since the trivial partition n is counted, so T(n,1) = 1.

%C This is also an irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists 1'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 A303300.

%e Triangle begins (rows 1..24).

%e 1;

%e 1;

%e 1;

%e 1;

%e 1;

%e 1;

%e 1;

%e 1, 1;

%e 1, 0;

%e 1, 1;

%e 1, 0;

%e 1, 1;

%e 1, 0;

%e 1, 1;

%e 1, 0;

%e 1, 1;

%e 1, 0;

%e 1, 1;

%e 1, 0;

%e 1, 1;

%e 1, 0, 1;

%e 1, 1, 0;

%e 1, 0, 0;

%e 1, 1, 1;

%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], so the 24th row of this triangle is [1, 1, 1].

%Y Row sums give A334948.

%Y Triangles of the same family where the parts differ by m are: A051731 (m=0), A237048 (m=1), A303300 (m=2), A330887 (m=3), A334460 (m=4), A334465 (m=5), this sequence (m=6).

%Y Cf. A000567, A334947, A334949, A334953.

%K nonn,tabf

%O 1

%A _Omar E. Pol_, May 27 2020