A317681 a(n) = smallest m such that sigma(m) = n*m/2.

1, 2, 6, 24, 120, 4320, 30240, 8910720, 14182439040, 17116004505600, 154345556085770649600, 170974031122008628879954060917200710847692800, 141310897947438348259849402738485523264343544818565120000

Offset

2,2

Comments

Interleaving of A007539 and A088912.

For even n, a(n) is a multiply perfect number; for odd n it is a hemiperfect number.

Note that 1 is the only number with abundancy 1, and 2 is the only number with abundancy 3/2 (in other words, 1 and 2 are solitary numbers; see A014567). For k >= 4 it is not known whether there are finitely many or infinitely many numbers with abundancy k/2. Also it is not known whether a(n) < a(n+1) always holds.

On the Riemann Hypothesis (RH), a(n) > exp(exp(n/(2*exp(gamma)))), where gamma = 0.5772156649... is the Euler-Mascheroni constant (A001620).

Links

Table of n, a(n) for n=2..14.

Achim Flammenkamp, The Multiply Perfect Numbers Page

Fred Helenius, Link to Glossary and Lists

G. P. Michon, Multiplicative functions: Abundancy = sigma(n)/n

G. P. Michon, Multiperfect and hemiperfect numbers

Walter Nissen, Abundancy: Some Resources

Formula

a(2n) = A007539(n), a(2n+1) = A088912(n), n > 0.

Example

a(7) = 4320 since sigma(4320) = 15120 = 7/2*4320 and 4320 is the smallest m such that sigma(m)/m = 7/2.

Mathematica

Nest[Append[#, Block[{m = #1[[-1]] + 1}, While[DivisorSigma[1, m] != #2 m/2, m++]; m]] & @@ {#, Length@ # + 2} &, {1}, 6] (* Michael De Vlieger, Aug 05 2018 *)

Prog

(PARI) for(n=2, 10, for(m=1, 10^12, if(sigma(m)/m==n/2, print1(m, ", "); break())))

Crossrefs

Cf. A007539, A088912, A246454.

Numbers with abundancy k/2: A000396 (k=4), A141643 (k=5), A005820 (k=6), A055153 (k=7), A027687 (k=8), A141645 (k=9), A046060 (k=10), A159271 (k=11), A046061 (k=12), A160678 (k=13).

Sequence in context: A037992 A335386 A114779 * A230363 A358500 A358494

Adjacent sequences: A317678 A317679 A317680 * A317682 A317683 A317684

Keyword

nonn,hard,more

Author

Jianing Song, Aug 04 2018

Status

approved