1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 2, 1, 1, 1, 3, 4, 3, 2, 1, 1, 1, 4, 5, 5, 3, 2, 1, 1, 1, 4, 7, 6, 5, 3, 2, 1, 1, 1, 5, 8, 9, 7, 5, 3, 2, 1, 1, 1, 5, 10, 11, 10, 7, 5, 3, 2, 1, 1, 1, 6, 12, 15, 13, 11, 7, 5, 3, 2, 1, 1, 1, 6, 14, 18, 18, 14, 11, 7, 5, 3, 2, 1, 1, 1, 7, 16, 23, 23, 20, 15, 11, 7, 5, 3, 2, 1, 1
Offset: 1
The triangle T(n,k) begins:
n\k 1 2 3 4 5 6 7 8 9 10 11 12 ...
1: 1
2: 1 1
3: 1 1 1
4: 1 2 1 1
5: 1 2 2 1 1
6: 1 3 3 2 1 1
7: 1 3 4 3 2 1 1
8: 1 4 5 5 3 2 1 1
9: 1 4 7 6 5 3 2 1 1
10: 1 5 8 9 7 5 3 2 1 1
11: 1 5 10 11 10 7 5 3 2 1 1
12: 1 6 12 15 13 11 7 5 3 2 1 1
... Reformatted and extended by _Wolfdieter Lang_, Dec 03 2012; additional extension by _Bob Selcoe_, Jun 09 2013
T(7,3) = 4 because we have [3,3,1], [3,2,2], [3,2,1,1] and [3,1,1,1,1], each having greatest part 3; or [5,1,1], [4,2,1], [3,3,1] and [3,2,2] each having 3 parts.
* Example from formula above: T(10,4) = 9 because T(6,4) + T(6,3) + T(6,2) + T(6,1) = 2 + 3 + 3 + 1 = 9.
* P(n) = P(n-1) + DT(n-1). P(n) = unordered partitions of n. (A000041) DT(n-1) = sum of diagonals beginning at T(n-1,1).
Example P(11) = 56, P(10) = 42, sum DT(10) = 1 + 4 + 5 + 3 + 1 = 14. - _Bob Selcoe_, Jun 09 2013
From _Omar E. Pol_, Nov 19 2019: (Start)
Illustration of initial terms: T(n,k) is also the number of vertical line segments in the k-th column of the n-th diagram, which represents the partitions of n:
.
1 1 1 1 1 1 1 2 1 1 1 2 2 1 1 1 3 3 2 1 1 1 3 4 3 2 1 1
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