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A286875
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If n = Product (p_j^k_j) then a(n) = Sum (k_j >= 2, p_j^k_j).
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2
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0, 0, 0, 4, 0, 0, 0, 8, 9, 0, 0, 4, 0, 0, 0, 16, 0, 9, 0, 4, 0, 0, 0, 8, 25, 0, 27, 4, 0, 0, 0, 32, 0, 0, 0, 13, 0, 0, 0, 8, 0, 0, 0, 4, 9, 0, 0, 16, 49, 25, 0, 4, 0, 27, 0, 8, 0, 0, 0, 4, 0, 0, 9, 64, 0, 0, 0, 4, 0, 0, 0, 17, 0, 0, 25, 4, 0, 0, 0, 16, 81, 0, 0, 4, 0, 0, 0, 8, 0, 9, 0, 4, 0, 0, 0, 32, 0, 49, 9, 29, 0, 0, 0, 8, 0, 0, 0, 31
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OFFSET
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1,4
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COMMENTS
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Sum of unitary, proper prime power divisors of n.
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LINKS
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FORMULA
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a(n) = Sum_{d|n, d = p^k, p prime, k >= 2, gcd(d, n/d) = 1} d.
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EXAMPLE
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a(360) = a(2^3*3^2*5) = 2^3 + 3^2 = 17.
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MATHEMATICA
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Table[DivisorSum[n, # &, CoprimeQ[#, n/#] && PrimePowerQ[#] && !PrimeQ[#] &], {n, 108}]
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PROG
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(Python)
from sympy import primefactors, isprime, gcd, divisors
def a(n): return sum(d for d in divisors(n) if gcd(d, n//d)==1 and len(primefactors(d))==1 and not isprime(d))
(PARI) A286875(n) = { my(f=factor(n)); for (i=1, #f~, if(f[i, 2] < 2, f[i, 1] = 0)); vecsum(vector(#f~, i, f[i, 1]^f[i, 2])); }; \\ Antti Karttunen, Oct 07 2017
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CROSSREFS
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Cf. A005117, A008475, A222416, A023888, A023889, A034448, A063956, A077610, A092261, A246547, A284117.
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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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