%I #25 Oct 09 2020 10:52:07
%S 1,1,1,1,1,2,1,0,1,2,1,0,1,2,1,0,1,2,1,0,3,1,2,0,1,0,0,1,2,3,1,0,0,1,
%T 2,0,1,0,3,1,2,0,1,0,0,1,2,3,1,0,0,4,1,2,0,0,1,0,3,0,1,2,0,0,1,0,0,4,
%U 1,2,3,0,1,0,0,0,1,2,0,0,1,0,3,4,1,2,0,0,1,0,0,0,1,2,3,0,1,0,0,4,1,2,0,0,5
%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 3, n >= 1, k >= 1, and the first element of column k is in the row that is the k-th pentagonal number (A000326).
%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 pentagonal number.
%F T(n,k) = k*A330887(n,k).
%e Triangle begins (rows 1..26):
%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, 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 1, 0, 0, 4;
%e 1, 2, 0, 0;
%e 1, 0, 3, 0;
%e 1, 2, 0, 0;
%e 1, 0, 0, 4;
%e ...
%e For n = 21 there are three partitions of 21 into consecutive parts that differ by 3, including 21 as a partition. They are [21], [12, 9] and [10, 7, 4]. The number of parts of these partitions are 1, 2 and 3 respectively, so the 21st row of the triangle is [1, 2, 3].
%p A330888 := proc(n,k)
%p k*A330887(n,k) ;
%p end proc:
%p for n from 1 to 40 do
%p for k from 1 do
%p if n>= A000325(k) then
%p printf("%d,",A330888(n,k)) ;
%p else
%p break;
%p end if;
%p end do:
%p printf("\n") ;
%p end do: # _R. J. Mathar_, Oct 02 2020
%Y Cf. A000326, A330887.
%Y Other triangles of the same family are A127093, A285914, A330466.
%K nonn,tabf
%O 1,6
%A _Omar E. Pol_, Apr 30 2020