OFFSET
1,3
COMMENTS
Let r = T(n,k) be a record in the sequence. The consecutive records "r" are the natural numbers A000027. Consider the first n rows; the triangle T(n,k) has the property that the columns, without the zeros, from k..1, are also the partitions of r in juxtaposed reverse-lexicographical order, so k is also A000041(r), the number of partitions of r. Note that a record r is always the final term of a row if such row contains 1’s. The number of positive integer a(1)..r is A006128(r). The sums a(1)..r is A066186(r). Here the set of positive integers in every row (from 1 to n) is called a “region” of r. The number of regions of r equals the number of partitions of r. If T(n,1) = 1 then the row n is formed by the smallest parts, in nondecreasing order, of all partitions of T(n,n).
LINKS
Robert Price, Table of n, a(n) for n = 1..196878, rows 1-627.
Omar E. Pol, Illustration of the seven regions of 5
EXAMPLE
Triangle begins:
1,
1, 2,
1, 1, 3,
0, 0, 0, 2,
1, 1, 1, 2, 4,
0, 0, 0, 0, 0, 3,
1, 1, 1, 1, 1, 2, 5,
0, 0, 0, 0, 0, 0, 0, 2,
0, 0, 0, 0, 0, 0, 0, 2, 4,
0, 0, 0, 0, 0, 0, 0, 0, 0, 3,
1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 6
...
The row n = 11 contains the 6th record in the sequence: a(66) = T(11,11) = 6, then consider the first 11 rows of triangle. Note that the columns, from k = 11..1, without the zeros, are also the 11 partitions of 6 in juxtaposed reverse-lexicographical order: [6], [3, 3], [4, 2], [2, 2, 2], [5, 1], [3, 2, 1], [4, 1, 1], [2, 2, 1, 1], [3, 1, 1, 1], [2, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1]. See A026792.
MATHEMATICA
"Table"], {_, _}][[All, 2]];
"Table"], {_, _}][[All, 2]];
f[n_] := Module[{v},
Reverse[PadRight[v, n]]];
Table[f[n], {n, PartitionsP[20]}] // Flatten (* Robert Price, Apr 26 2020 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Omar E. Pol, Aug 08 2011
STATUS
approved