A348148 Numbers k for which sigma(k)/k = 32/9.

3780, 66960, 167400, 406224, 1097280, 6656832, 13035330, 29410290, 4529295360, 27477725184, 88071903612, 1159632322560, 7035102756864, 18554223329280, 22385029489560, 54934276752360, 112562288197632, 125356165141536, 307631949813216

Offset: 1

Author

Timothy L. Tiffin, Oct 02 2021

Keywords

• nonn

Comments

This sequence will contain terms of the form 135*P and 819*Q, where P is a perfect number (A000396) not divisible by 3 or 5, and Q is a perfect number not divisible by 3, 7, or 13. Proof: sigma(135*P)/(135*P) = sigma(135)*sigma(P)/(135*P) = 240*(2*P)/(135*P) = 32/9 and sigma(819*Q)/(819*Q) = sigma(819)*sigma(Q)/(819*Q) = 1456*(2*Q)/(819*P) = 32/9. QED

Terms ending in "4", "32", or "80" and some terms ending in "60" will have one of these forms:

a( 1) = 3780 = 135* 28 = 135*A000396(2)

a( 2) = 66960 = 135* 496 = 135*A000396(3)

a( 4) = 406224 = 819* 496 = 819*A000396(3)

a( 5) = 1097280 = 135* 8128 = 135*A000396(4)

a( 6) = 6656832 = 819* 8128 = 819*A000396(4)

a( 9) = 4529295360 = 135* 33550336 = 135*A000396(5)

a(10) = 27477725184 = 819* 33550336 = 819*A000396(5)

a(12) = 1159632322560 = 135* 8589869056 = 135*A000396(6)

a(13) = 7035102756864 = 819* 8589869056 = 819*A000396(6)

a(14) = 18554223329280 = 135*137438691328 = 135*A000396(7)

a(17) = 112562288197632 = 819*137438691328 = 819*A000396(7).

Examples

			167400 is a term, since sigma(167400)/167400 = 595200/167400 = 32/9.
		

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, A211680, A212610.

Subsequence of A005101 and A218416.

Programs

Mathematica

Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/9 &]
Do[If[DivisorSigma[1, k]/k == 32/9, Print[k]], {k, 5*10^8}]