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A194809
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Imbalance of the sum of largest parts of all partitions of n.
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2
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0, -2, 1, -5, 3, -12, 7, -25, 17, -47, 36, -88, 69, -155, 133, -262, 240, -439, 415, -717, 705, -1142, 1165, -1803, 1874, -2797, 2975, -4276, 4632, -6478, 7094, -9698, 10741, -14355, 16059, -21079, 23719, -30670, 34716, -44243, 50315, -63372
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OFFSET
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1,2
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COMMENTS
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Consider the three-dimensional structure of the shell model of partitions version "tree". Note that only the larges parts > 1 produce the imbalance. Note that every column where is located a largest part contains largest parts of the same size, thesame as a periodic table (see example). For more information see A135010.
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LINKS
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FORMULA
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a(n) = Sum_{k=2..n} (-1)^(k-1)*A138137(k), n >= 2.
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EXAMPLE
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For n = 6 the illustration of the shell model with 6 shells shows an imbalance of largest parts (see below):
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Partitions Tree Table 1.0
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6 6 6 . . . . .
3+3 3 3 . . 3 . .
4+2 4 4 . . . 2 .
2+2+2 2 2 . 2 . 2 .
5+1 1 5 5 . . . . 1
3+2+1 1 3 3 . . 2 . 1
4+1+1 4 1 4 . . . 1 1
2+2+1+1 2 1 2 . 2 . 1 1
3+1+1+1 1 3 3 . . 1 1 1
2+1+1+1+1 2 1 2 . 1 1 1 1
1+1+1+1+1+1 1 1 1 1 1 1 1
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The sum of largest parts > 1 on the left hand side is 23 and the sum of largest parts > 1 on the right hand side is 11, so a(6) = -23 + 11 = -12. On the other hand for n = 6 we have that 0 together with the first n-1 terms > 1 of A138137 are 0, 2, 3, 6, 8, 15 so a(6) = 0-2+3-6+8-15 = -12.
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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