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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,5
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COMMENTS
| 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
| 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
| a(15)=3 because we have [15],[9,6] and [8,5,2].
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MAPLE
| 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
| Cf. A000005, A001227, A038548.
Sequence in context: A003640 A107459 A087976 * A033831 A033105 A106703
Adjacent sequences: A117274 A117275 A117276 * A117278 A117279 A117280
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KEYWORD
| nonn
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AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 07 2006
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