OFFSET
0,4
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).
LINKS
Alois P. Heinz, Rows n = 0..140, flattened (first 22 rows from Clark Kimberling)
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
AUTHOR
Clark Kimberling, Mar 28 2014
STATUS
approved