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A091051
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Sum of divisors of n that are perfect powers.
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5
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1, 1, 1, 5, 1, 1, 1, 13, 10, 1, 1, 5, 1, 1, 1, 29, 1, 10, 1, 5, 1, 1, 1, 13, 26, 1, 37, 5, 1, 1, 1, 61, 1, 1, 1, 50, 1, 1, 1, 13, 1, 1, 1, 5, 10, 1, 1, 29, 50, 26, 1, 5, 1, 37, 1, 13, 1, 1, 1, 5, 1, 1, 10, 125, 1, 1, 1, 5, 1, 1, 1, 58, 1, 1, 26, 5, 1, 1, 1, 29, 118, 1, 1, 5, 1, 1, 1, 13, 1, 10
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OFFSET
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1,4
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COMMENTS
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a(p^k) = 1+(p^2)*(p^(k-1)-1)/(p-1) for p prime, k>0.
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LINKS
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FORMULA
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G.f.: Sum_{k=i^j, i>=1, j>=2, excluding duplicates} k*x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 20 2017
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EXAMPLE
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Divisors of n=108: {1,2,3,4,6,9,12,18,27,36,54,108}, a(108) = 1^2 + 2^2 + 3^2 + 3^3 + 6^2 = 1+4+9+27+36 = 77.
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MATHEMATICA
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a[n_] := DivisorSum[n, #*Boole[# == 1 || GCD @@ FactorInteger[#][[All, 2]] >1]&]; Array[a, 90] (* Jean-François Alcover, May 09 2017 *)
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PROG
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(PARI) a(n) = sumdiv(n, d, d*((d==1) || ispower(d))); \\ Michel Marcus, Oct 02 2014
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CROSSREFS
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Differs from A183097 for the first time at n=72.
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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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