A239830 Triangular array: T(n,k) = number of partitions of 2n that have alternating sum 2k, with T(0,0) = 1 for convenience.

1, 1, 1, 2, 2, 1, 3, 5, 2, 1, 5, 9, 5, 2, 1, 7, 17, 10, 5, 2, 1, 11, 28, 20, 10, 5, 2, 1, 15, 47, 35, 20, 10, 5, 2, 1, 22, 73, 62, 36, 20, 10, 5, 2, 1, 30, 114, 102, 65, 36, 20, 10, 5, 2, 1, 42, 170, 167, 109, 65, 36, 20, 10, 5, 2, 1, 56, 253, 262, 182, 110

Offset: 0

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) = A000041(n) for n >= 0, which is the number of partitions of 2n having AS(p) = 0, for n >= 1. Columns 2 and 3 are essentially A000567 and A000710, and the limiting column (after deleting initial 0's), A000712. The sum of numbers in row n is A000041(2n). The corresponding array for partitions into distinct parts is given by A152146 (defined as the number of unrestricted partitions of 2n into 2k even parts).

Examples

			First nine rows:
1
1 ... 1
2 ... 2 ... 1
3 ... 5 ... 2 ... 1
5 ... 9 ... 5 ... 2 ... 1
7 ... 17 .. 10 .. 5 ... 2 ... 1
11 .. 28 .. 20 .. 10 .. 5 ... 2 ... 1
15 .. 47 .. 35 .. 20 .. 10 .. 5 ... 2 ... 1
22 .. 73 .. 62 .. 36 .. 20 .. 10 .. 5 ... 2 ... 1
The partitions of 6 are 6, 51, 42, 411, 33, 321, 3111, 222, 2211, 21111, 111111, with respective alternating sums 6, 4, 2, 4, 0, 2, 2, 2, 0, 2, 0, so that row 3 (counting the top row as row 0) of the array is 3 .. 5 .. 2 .. 1.
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened (first 22 rows from Clark Kimberling)

Crossrefs

Cf. A239829, A239830, A239833.

Programs

Maple

b:= proc(n, i, t) option remember; `if`(n=0, 1, `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=0..n))(b(2*n$2, 1)):
seq(T(n), n=0..14);  # Alois P. Heinz, Mar 30 2014

Mathematica

z = 16; 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], 2 k]; t[0, 0] = 1; u = Table[t[n, k], {n, 0, z}, {k, 0, n}]
TableForm[u]  (* A239830, array *)
Flatten[u]    (* A239830, sequence *)
(* Peter J. C. Moses, Mar 21 2014 *)