proposed

approved

editing

proposed

allocated for Clark KimberlingTriangular array: t(n,k) = number of partitions p = {x(1) >= x(2) >= ... >= x(k)} such that min(x(j) - x(j-1)) = k.

1, 1, 1, 3, 0, 1, 4, 1, 0, 1, 7, 1, 1, 0, 1, 10, 2, 0, 1, 0, 1, 16, 2, 1, 0, 1, 0, 1, 22, 3, 1, 1, 0, 1, 0, 1, 32, 4, 2, 0, 1, 0, 1, 0, 1, 44, 5, 2, 1, 0, 1, 0, 1, 0, 1, 62, 6, 3, 1, 1, 0, 1, 0, 1, 0, 1, 83, 8, 3, 2, 0, 1, 0, 1, 0, 1, 0, 1, 113, 10, 4, 2, 1

1,4

The first two columns are essentially A047967 and A238708. Counting the top row as row 2, the sum of numbers in row n is A000041(n) - 1.

Clark Kimberling, <a href="/A238709/b238709.txt">Table of n, a(n) for n = 1..400</a>

row 2: 1

row 3: 1 ... 1

row 4: 3 ... 0 ... 1

row 5: 4 ... 1 ... 0 ... 1

row 6: 7 ... 1 ... 1 ... 0 ... 1

row 7: 10 .. 2 ... 0 ... 1 ... 0 ... 1

row 8: 16 .. 2 ... 1 ... 0 ... 1 ... 0 ... 1

row 9: 22 .. 3 ... 1 ... 1 ... 0 ... 1 ... 0 ... 1

Let m = min(x(j) - x(j-1)); then for row 5, the 4 partitions with m = 0 are 311, 221, 2111, 11111; the 1 partition with m = 1 is 32, and the 1 partition with m = 3 is 41.

z = 25; p[n_, k_] := p[n, k] = IntegerPartitions[n][[k]]; m[n_, k_] := m[n, k] = Min[-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]

Cf. A238710, A238708.

allocated

nonn,tabl,easy

Clark Kimberling, Mar 03 2014

approved

editing

allocated for Clark Kimberling

allocated

approved