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A239830
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Triangular array: T(n,k) = number of partitions of 2n that have alternating sum 2k, with T(0,0) = 1 for convenience.
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37
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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
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OFFSET
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0,4
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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.
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MAPLE
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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)):
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MATHEMATICA
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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}]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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