A347203 Numbers k for which sigma(k)/k = 32/13.

546, 45136, 739648, 5752422, 3053080576, 781678084096, 12506920910848, 209831713740735643648, 241919495232854688763577028051799638016

Offset: 1

Author

Timothy L. Tiffin, Aug 22 2021

Keywords

• nonn
• more

Comments

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.

Examples

			5752422 is a term, since sigma(5752422)/5752422 = 14159808/5752422 = 32/13.
		

Links

• G. P. Michon, Multiperfect Numbers and Hemiperfect Numbers
• Walter Nissen, Abundancy: Some Resources (preliminary version 4)
• Walter Nissen, Primitive Friendly Pairs with friends < 2^34 with denom < 20000

Crossrefs

Cf. A000203, A000396.

Subsequence of A005101.

Programs

Mathematica

Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/13 &]
Do[If[DivisorSigma[1, k]/k == 32/13, Print[k]], {k, 5*10^8}]

Extensions

a(8)-a(9) from Michel Marcus, Aug 23 2021