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A347203
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Numbers k for which sigma(k)/k = 32/13.
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0
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OFFSET
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1,1
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COMMENTS
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This sequence will contain terms of the form 91*P, where P is a perfect number (A000396) not divisible by 7 or 13. Proof: sigma(91*P)/(91*P) = sigma(91)*sigma(P)/(91*P) = 112*(2*P)/(91*P) = 32/13. QED.
Terms ending in "6" or "48" have this form. Example: a(n) = 91*A000396(n) for n = 1, 5, 6, 7, 8, 9 and a(n) = 91*A000396(n+1) for n = 2, 3.
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LINKS
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EXAMPLE
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5752422 is a term, since sigma(5752422)/5752422 = 14159808/5752422 = 32/13.
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MATHEMATICA
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Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/13 &]
Do[If[DivisorSigma[1, k]/k == 32/13, Print[k]], {k, 5*10^8}]
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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