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A117277
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Number of partitions of n whose consecutive parts differ by 3.
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13
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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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COMMENTS
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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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LINKS
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FORMULA
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G.f.: Sum_{k>=1} x^((3*k^2-k)/2)/(1-x^k). In general, the generating function for the number of partitions in which consecutive parts differ by d is Sum_{k>=1} x^(k*(d*k-d+2)/2)/(1-x^k). 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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PROG
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(PARI) seq(N, d)=my(x='x+O('x^N)); Vec(sum(k=1, N, x^(k*(d*k-d+2)/2)/(1-x^k)));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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