A238710 Triangular array: t(n,k) = number of partitions p = {x(1) >= x(2) >= ... >= x(k)} such that max(x(j) - x(j-1)) = k.

1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 3, 3, 2, 1, 1, 1, 6, 3, 2, 1, 1, 3, 6, 6, 2, 2, 1, 1, 2, 10, 6, 5, 2, 2, 1, 1, 3, 11, 11, 6, 4, 2, 2, 1, 1, 1, 16, 13, 10, 5, 4, 2, 2, 1, 1, 5, 17, 19, 12, 9, 4, 4, 2, 2, 1, 1, 1, 24, 24, 18, 11, 8, 4, 4, 2, 2, 1, 1, 3, 27, 34

Offset: 1

Author

Clark Kimberling, Mar 03 2014

Keywords

• nonn
• tabl
• easy

Comments

The first two columns are essentially A032741 and A237665. Counting the top row as row 2, the sum of numbers in row n is A000041(n) - 1.

Examples

			row 2:  1
row 3:  1 ... 1
row 4:  2 ... 1 ... 1
row 5:  1 ... 3 ... 1 ... 1
row 6:  3 ... 3 ... 2 ... 1 ... 1
row 7:  1 ... 6 ... 3 ... 2 ... 1 ... 1
row 8:  3 ... 6 ... 6 ... 2 ... 2 ... 1 ... 1
row 9:  2 ... 10 .. 6 ... 5 ... 2 ... 2 ... 1 ... 1
Let m = max(x(j) - x(j-1)); then for row 5, the 1 partition with m = 0 is 11111; the 3 partitions with m = 1 are 32, 221, 2111; the 1 partition with m = 2 is 311, and the 1 partition with m = 3 is 41.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..400

Crossrefs

Cf. A238710, A238709, A238353.

Programs

Mathematica

z = 25; p[n_, k_] := p[n, k] = IntegerPartitions[n][[k]]; m[n_, k_] := m[n, k] = Max[-Differences[p[n, k]]]; c[n_] := Table[m[n, h], {h, 1, PartitionsP[n]}]; v = Table[Count[c[n], h], {n, 2, z}, {h, 0, n - 2}]; Flatten[v]
TableForm[v]