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Search: id:A117277
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| A117277 |
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Number of partitions of n whose consecutive parts differ by 3. |
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+0
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| 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 3, 1, 2, 3, 2, 1, 3, 2, 3, 2, 2, 2, 3, 2, 2, 3, 2, 1, 4, 2, 2, 2, 2, 3, 4, 1, 2, 3, 3, 1, 4, 2, 2, 3, 2, 2, 4, 1, 3, 3, 2, 1, 4, 4, 2, 2, 2, 2, 5, 1, 3, 3, 2, 2, 4, 2, 2, 3, 3, 2, 4, 1, 2, 4, 3, 2, 4, 2, 3, 2, 2, 3, 4, 3, 2, 3, 2, 1, 6
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OFFSET
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1,5
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COMMENT
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Also number of partitions of n such that if k is the largest part, then each of the parts 1,2,...,k-1 occurs exactly 3 times. Example: a(15)=3 because we have [3,3,2,2,2,1,1,1],[2,2,2,2,2,2,1,1,1], and [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1].
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FORMULA
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G.f.=sum(x^((3k^2-k)/2)/(1-x^k), k=1..infinity). In general, the generating function for the number of partitions in which consecutive parts differ by d is sum(x^(k(dk-d+2)/2)/(1-x^k), k=1..infinity). For d=0,1, and 2 one obtains A000005,A001227, and A038548, respectively.
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EXAMPLE
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a(15)=3 because we have [15],[9,6], and [8,5,2].
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MAPLE
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g:=sum(x^((3*k^2-k)/2)/(1-x^k), k=1..10): gser:=series(g, x=0, 140): seq(coeff(gser, x^n), n=1..135);
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CROSSREFS
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Cf. A000005, A001227, A038548.
Sequence in context: A003640 A107459 A087976 this_sequence A033831 A033105 A106703
Adjacent sequences: A117274 A117275 A117276 this_sequence A117278 A117279 A117280
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KEYWORD
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nonn
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 07 2006
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