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A194260
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A194259(n) - n, where A194259(n) is the number of distinct prime factors of p(1)*p(2)*...*p(n) and p(n) is the n-th partition number.
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5
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-1, -1, -1, -1, -1, -1, -2, -3, -4, -5, -6, -7, -7, -8, -9, -10, -11, -12, -13, -13, -14, -14, -14, -15, -15, -15, -15, -15, -15, -15, -15, -15, -16, -16, -16, -16, -16, -17, -18, -18, -18, -18, -18, -17, -17, -16, -16, -16, -16, -16, -16, -16, -16, -15, -15, -14, -14, -14, -14, -13, -13, -13, -12, -12, -12, -12, -11, -11, -10, -10, -10, -10, -9, -9, -9, -9, -9, -8, -7, -7, -7, -8, -8, -8, -8, -7, -7, -7, -7, -6, -5, -4, -4, -4, -3, -3, -4, -4, -4, -4, -4, -3, -3, -3, -3, -3, -3, -3, -3, -2, -2, -2, -2, -2, 0, 1
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OFFSET
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1,7
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COMMENTS
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Schinzel and Wirsing proved that a(n) > C*log n - n, for any positive constant C < 1/log 2 and all large n. In fact, it appears that a(n) > 0 for all n > 115.
It also appears that a(n) >= a(n-1), for all n > 97, so that some prime factor of p(n) does not divide p(1)*p(2)*...*p(n-1).
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LINKS
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FORMULA
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EXAMPLE
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p(1)*p(2)*...*p(8) = 1*2*3*5*7*11*15*22 = 2^2 * 3^2 * 5^2 * 7 * 11^2, so a(8) = 5 - 8 = -3.
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MAPLE
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with(combinat): with(numtheory):
b:= proc(n) option remember;
`if`(n=1, {}, b(n-1) union factorset(numbpart(n)))
end:
a:= n-> nops(b(n)) -n:
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MATHEMATICA
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a[n_] := PrimeNu[Product[PartitionsP[k], {k, 1, n}]] - n; Table[a[n], {n, 1, 116}] (* Jean-François Alcover, Jan 28 2014 *)
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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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