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A332889
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a(n) = number of strict partition numbers >1 that are proper divisors of the n-th strict partition number.
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1
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0, 0, 0, 1, 0, 2, 2, 2, 4, 2, 3, 1, 1, 3, 1, 1, 5, 4, 3, 0, 3, 1, 1, 3, 8, 3, 5, 3, 4, 6, 5, 6, 3, 2, 7, 10, 1, 1, 9, 2, 4, 3, 7, 11, 3, 6, 9, 1, 0, 1, 9, 3, 3, 2, 1, 6, 11, 8, 2, 1, 7, 2, 6, 2, 4, 12, 3, 0, 4, 8, 4, 4, 1, 7, 0, 1, 9, 7, 5, 5, 1, 1, 6, 5, 4
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OFFSET
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3,6
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COMMENTS
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Let p(n) = number of strict partitions of n. Then p(11) = 12, which is divisible by these 6 strict partition numbers: p(2) = 1, p(3) = 2, p(5) = 3, p(6) = 4, p(8) = 6, and p(11) = 12; discounting 1 and 12 leaves a(11) = 4 divisors.
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LINKS
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FORMULA
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MATHEMATICA
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p[n_] := PartitionsQ[n]; t[n_] := Table[p[k], {k, 0, n}]
-2+Table[Length[Intersection[t[n], Divisors[p[n]]]], {n, 3, 130}]
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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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