|
%I #15 Oct 02 2020 09:58:24
%S 1,1,1,1,1,1,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,0,3,1,2,0,
%T 1,0,0,1,2,3,1,0,0,1,2,0,1,0,3,1,2,0,1,0,0,1,2,3,1,0,0,1,2,0,1,0,3,1,
%U 2,0,1,0,0,1,2,3,1,0,0,4,1,2,0,0,1,0,3,0,1,2,0,0,1,0,0,4,1,2,3,0,1,0,0,0,1
%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 5, n >= 1, k >= 1, and the first element of column k is in the row that is the k-th heptagonal number (A000566).
%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 heptagonal 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*A334465(n,k).
%e Triangle begins (rows 1..27):
%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, 3;
%e 1, 2, 0;
%e 1, 0, 0;
%e 1, 2, 3;
%e 1, 0, 0;
%e 1, 2, 0;
%e 1, 0, 3;
%e 1, 2, 0;
%e 1, 0, 0;
%e 1, 2, 3;
%e ...
%e For n = 27 there are three partitions of 27 into consecutive parts that differ by 5, including 27 as a valid partition. They are [27], [16, 11] and [14, 9, 4]. The number of parts of these partitions are 1, 2, 3 respectively, so the 27th row of the triangle is [1, 2, 3].
%p A334540 := proc(n,k)
%p k*A334465(n,k) ;
%p end proc: # _R. J. Mathar_, Oct 02 2020
%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), this sequence (d=5).
%Y Cf. A000566, A334465.
%K nonn,tabf
%O 1,8
%A _Omar E. Pol_, May 05 2020
|
