|
%I #29 May 05 2020 05:32:16
%S 1,1,1,1,2,1,0,1,2,1,0,1,2,1,0,3,1,2,0,1,0,0,1,2,3,1,0,0,1,2,0,1,0,3,
%T 1,2,0,4,1,0,0,0,1,2,3,0,1,0,0,0,1,2,0,4,1,0,3,0,1,2,0,0,1,0,0,0,1,2,
%U 3,4,1,0,0,0,5,1,2,0,0,0,1,0,3,0,0,1,2,0,4,0,1,0,0,0,0,1,2,3,0,5,1,0,0,0,0
%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 2, n >= 1, k >= 1, and the first element of column k is in row k^2.
%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 row k^2.
%C Conjecture: row sums give A066839.
%F T(n,k) = k*A303300(n,k).
%e Triangle begins (rows 1..25):
%e 1;
%e 1;
%e 1;
%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, 4;
%e 1, 0, 0, 0;
%e 1, 2, 3, 0;
%e 1, 0, 0, 0;
%e 1, 2, 0, 4;
%e 1, 0, 3, 0;
%e 1, 2, 0, 0;
%e 1, 0, 0, 0;
%e 1, 2, 3, 4;
%e 1, 0, 0, 0, 5;
%e ...
%e For n = 16 there are three partitions of 16 into consecutive parts that differ by 2, including 16 as a partition. They are [16], [9, 7] and [7, 5, 3, 1]. The number of parts of these partitions are 1, 2 and 4 respectively, so the 16th row of the triangle is [1, 2, 0, 4].
%Y Cf. A000290, A066839, A237048, A303300.
%Y Other triangles of the same family are A127093 and A285914.
%K nonn,tabf
%O 1,5
%A _Omar E. Pol_, Apr 30 2020
|
