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A347261
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Numbers k for which sigma(k)/k = 32/11.
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1
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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 33*P, where P is a perfect number (A000396) not divisible by 3 or 11. Proof: sigma(33*P)/(33*P) = sigma(33)*sigma(P)/(33*P) = 48*(2*P)/(33*P) = 32/11. QED
Terms ending in "24" or "x8" (where x is an even digit) have this form. Example: a(1) = 33*A000396(2), a(3) = 33*A000396(3), and a(n) = 33*A000396(n-1) for 5, 6, 7, 8.
Conjecture: a(n) = 33*A341623(n) for n >= 1. Motivation: If no term of A341623 is divisible by 11 (which appears to be the case), then sigma(33*A341623(n))/(33*A341623(n)) = sigma(11)*sigma(3*A341623(n))/(33*A341623(n)) = 12*(8*A341623(n))/(33*A341623(n)) = 32/11. Does this sequence, though, contain any additional terms that are not generated by A341623?
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LINKS
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EXAMPLE
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2970 is a term, since sigma(2970)/2970 = 8640/2970 = 32/11.
18018 is a term, since sigma(18018)/18018 = 52416/18018 = 32/11.
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MATHEMATICA
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Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/11 &]
Do[If[DivisorSigma[1, k]/k == 32/11, 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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