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A351539
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Number of odd prime factors p of n such that p^(1+valuation(n,p)) divides sigma(n).
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3
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
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OFFSET
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1,2235
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LINKS
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FORMULA
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EXAMPLE
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For n = 2235 = 3 * 5 * 149, sigma(2235) = 3600 = 2^4 * 3^2 * 5^2, and thus for two of the three odd prime factors of 2235, a higher power of the same prime divides sigma(2235), and therefore a(2235) = 2. This is also the least n for which a(n) > 1.
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MATHEMATICA
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{0}~Join~Table[Function[m, Count[DeleteCases[FactorInteger[n][[All, 1]], 2], _?(Mod[m, #^(1 + IntegerExponent[n, #])] == 0 &)]][DivisorSigma[1, n]], {n, 2, 120}] (* Michael De Vlieger, Feb 16 2022 *)
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PROG
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(PARI) A351539(n) = { my(f=factor(n), s=sigma(n)); sum(k=1, #f~, (f[k, 1]%2)&&(0==(s%(f[k, 1]^(1+f[k, 2]))))); };
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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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