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A245525
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Unique integer r with -prime(n)/2 < r <= prime(n)/2 such that p(n) == r (mod prime(n)), where p(.) is the partition function given by A000041.
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5
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1, -1, -2, -2, -4, -2, -2, 3, 7, 13, -6, 3, 19, 6, -12, 19, 2, 19, 21, -12, -11, -25, 10, -27, 18, 12, 23, -27, -13, -46, -16, -35, 5, -61, -17, 8, -29, -65, -44, -30, 12, -40, 40, -95, 90, 88, 53, 93, 97, -42, -47, 47, 2, 117, -16, 34, 27, 51, -11, 108, -24, 115, -29, 30, -32, -90, -87, 141, 24, 131, -166, -115, -96, -111, 84, -191, 163, -156, 115, 78
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OFFSET
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1,3
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COMMENTS
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Conjecture: a(n) is always nonzero, i.e., prime(n) never divides the partition number p(n).
This conjecture does not hold with the smallest counterexample being n=1119414 (cf. A245662). - Max Alekseyev, Jul 27 2014
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LINKS
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FORMULA
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EXAMPLE
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a(20) = -12 since p(20) = 627 == -12 (mod prime(20)=71).
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MATHEMATICA
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rMod[m_, n_]:=Mod[m, n, -(n-1)/2]
a[n_]:=rMod[PartitionsP[n], Prime[n]]
Table[a[n], {n, 1, 80}]
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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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