OFFSET
1,1
COMMENTS
There are no other terms up to 1.2*10^14.
If P (A000396) is an even perfect number, then 4*P is in the sequence. Are there any others?
If there are no terms of another form, the sequence goes on with 9223372032559808512 = 2^32 * A000668(8), 10633823966279326978618770463815368704 = 2^62 * A000668(9), 766247770432944429179173512337214552523989286192676864 = 2^90 * A000668(10), ... - Michel Lagneau, Jan 21 2015
Conjecture: let s0 be the sum of the inverses of the even divisors of a number n and s1 the sum of the inverses of the odd divisors of n; then n is in the sequence iff 2*s0-s1 = 1. - Michel Lagneau, Jan 21 2015
Ligh & Wall proved that 2^(p+1)*(2^p-1) is a term if p and 2^p-1 are primes, and that all the nonunitary perfect numbers below 10^6 are of this form. - Amiram Eldar, Sep 27 2018
If k is in the sequence and k = 2^m*p^a then k is of the form 4*P for an even perfect P. See the link to MathOverflow. - Joshua Zelinsky, Mar 07 2024
LINKS
Steve Ligh and Charles R. Wall, Functions of Nonunitary Divisors, Fibonacci Quarterly, Vol. 25 (1987), pp. 333-338.
EXAMPLE
The sum of the nonunitary divisors of 24 is 2 + 4 + 6 + 12 = 24.
MATHEMATICA
nusigma[ n_ ] := DivisorSigma[ 1, n ]-Times@@(1+Power@@#&/@FactorInteger[ n ]); For[ n=1, True, n++, If[ nusigma[ n ]==n, Print[ n ] ] ]
Do[s0=0; s1=0; Do[d=Divisors[n][[i]]; If[Mod[d, 2]==0, s0=s0+1/d, s1=s1+1/d], {i, 1, Length[Divisors[n]]}]; If[2*s0-s1==1, Print[n]], {n, 2, 10^9, 2}] (* Michel Lagneau, Jan 21 2015 *)
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Dean Hickerson, Sep 25 2001
STATUS
approved