A239302 Triangular array: T(n,k) = number of partitions x(1) > x(2) > ... > x(k) of n+2 such that x(1) = x(2) + k, for n >= 1.

1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 2, 1, 1, 1, 1, 0, 1, 2, 2, 1, 1, 1, 1, 0, 1, 2, 2, 2, 1, 1, 1, 1, 0, 1, 3, 2, 2, 2, 1, 1, 1, 1, 0, 1, 3, 3, 2, 2, 2, 1, 1, 1, 1, 0, 1, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 0, 1, 5, 4, 3, 3, 2, 2, 2, 1

Offset

1,22

Comments

The columns are identical, and the limit of the reversal of rows is A087897(n+3).

Sum of row n equals A111133(n+3).

Links

Clark Kimberling, Table of n, a(n) for n = 1..1000

Example

First 17 rows:
1
0 1
1 0 1
1 1 0 1
1 1 1 0 1
1 1 1 1 0 1
2 1 1 1 1 0 1
2 2 1 1 1 1 0 1
2 2 2 1 1 1 1 0 1
3 2 2 2 1 1 1 1 0 1
3 3 2 2 2 1 1 1 1 0 1
4 3 3 2 2 2 1 1 1 1 0 1
5 4 3 3 2 2 2 1 1 1 1 0 1
5 5 4 3 3 2 2 2 1 1 1 1 0 1
6 5 5 4 3 3 2 2 2 1 1 1 1 0 1
8 6 5 5 4 3 3 2 2 2 1 1 1 1 0 1
8 8 6 5 5 4 3 3 2 2 2 1 1 1 1 0 1
To account for row 7, start with the strict partitions (A000009) of 9 that have more than one part:  81, 72, 63, 621, 54, 531, 432.  Next, form (part 1) - (part 2) for each of those partitions, getting 7, 5, 3, 4, 1, 2, 1; finally, note that the numbers of occurrences of 1,2,3,4,5,6,7, respectively, are 2,1,1,1,1,0,1.

Mathematica

z = 25; d[n_] := d[n] = Rest[Select[IntegerPartitions[n], DeleteDuplicates[#] == # &]]; t[n_] := t[n] = Table[d[n][[k, 1]] - d[n][[k, 2]], {k, 1, -1 + PartitionsQ[n]}]; u = Table[Count[t[n], j], {n, 3, z}, {j, 1, n - 2}]; TableForm[u] (* A239302 as an array *)
v = Flatten[u]  (* A239302 as a sequence *)

Crossrefs

Cf. A000009, A111133, A087897.

Sequence in context: A087116 A033264 A258045 * A256983 A330720 A080234

Adjacent sequences: A239299 A239300 A239301 * A239303 A239304 A239305

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Mar 14 2014

Status

approved