A239829 Triangular array: T(n,k) = number of partitions of 2n - 1 that have alternating sum 2k - 1.

1, 2, 1, 4, 2, 1, 7, 5, 2, 1, 12, 10, 5, 2, 1, 19, 19, 10, 5, 2, 1, 30, 33, 20, 10, 5, 2, 1, 45, 57, 36, 20, 10, 5, 2, 1, 67, 92, 64, 36, 20, 10, 5, 2, 1, 97, 147, 107, 65, 36, 20, 10, 5, 2, 1, 139, 227, 177, 110, 65, 36, 20, 10, 5, 2, 1, 195, 345, 282, 184

Offset: 1

Author

Clark Kimberling, Mar 28 2014

Keywords

• nonn
• tabl
• easy

Comments

Suppose that p, with parts x(1) >= x(2) >= ... >= x(k), is a partition of n. Define AS(p), the alternating sum of p, by x(1) - x(2) + x(3) - ... + ((-1)^(k-1))*x(k); note that AS(p) has the same parity as n. Column 1 is given by T(n,1) = (number of partitions of 2n-1 having AS(p) = 1) = A000070(n) for n >= 1. Columns 2 and 3 are essentially A000098 and A103924, and the limiting column (after deleting initial 0's), A000712. The sum of numbers in row n is A000041(2n-1). The corresponding array for partitions into distinct parts is given by A152157 (defined as the number of partitions of 2n+1 into 2k+1 odd parts).

Examples

			First nine rows:
1
2 ... 1
4 ... 2 ... 1
7 ... 5 ... 2 ... 1
12 .. 10 .. 5 ... 2 ... 1
19 .. 19 .. 10 .. 5 ... 2 ... 1
30 .. 33 .. 20 .. 10 .. 5 ... 2 ... 1
45 .. 57 .. 36 .. 20 .. 10 .. 5 ... 2 ... 1
67 .. 92 .. 64 .. 36 .. 20 .. 10 .. 5 ... 2 ... 1
The partitions of 5 are 5, 41, 32, 311, 221, 2111, 11111, with respective alternating sums 5, 3, 1, 3, 1, 1, 1, so that row 2 of the array is 4 .. 2 .. 1.
		

Links

• Alois P. Heinz, Rows n = 1..141, flattened (first 20 rows from Clark Kimberling)

Crossrefs

Cf. A239830, A239832, A239833.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n=0, x^(1/2), `if`(i<1, 0,
      expand(b(n, i-1, t)+`if`(i>n, 0, b(n-i, i, -t)*x^((t*i)/2)))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(2*n-1$2, 1)):
seq(T(n), n=1..14);  # Alois P. Heinz, Mar 30 2014

Mathematica

z = 15; s[w_] := s[w] = Total[Take[#, ;; ;; 2]] - Total[Take[Rest[#], ;; ;; 2]] &[w]; c[n_] := c[n] = Table[s[IntegerPartitions[n][[k]]], {k, 1, PartitionsP[n]}]; t[n_, k_] := Count[c[2 n - 1], 2 k - 1]; u = Table[t[n, k], {n, 1, z}, {k, 1, n}]
TableForm[u]  (* A239829, array *)
Flatten[u]    (* A239829, sequence *)
(* Peter J. C. Moses, Mar 21 2014 *)
b[n_, i_, t_] := b[n, i, t] = If[n == 0, x^(1/2), If[i<1, 0, Expand[b[n, i-1, t] + If[i>n, 0, b[n-i, i, -t]*x^((t*i)/2)]]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[2n-1, 2n-1, 1]]; Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, Aug 27 2016, after Alois P. Heinz *)