OFFSET
1,1
COMMENTS
Includes 2*p for p in A135535. - Robert Israel, May 13 2025
From David A. Corneth, May 13 2025: (Start)
If a(n) = m*p where p is the largest prime divisor and has multiplicity 1 and s = sigma(m) then p = (4^k - s) / (s - m). Using this, a(44), a(45) and a(46) are at most 4971704751572, 44088037271892 and 44358570286896 respectively.
If a(n) is odd then a(n) is a perfect square. Proof: Suppose a(n) is not a perfect square. Then sigma(a(n)) is even and so sigma(a(n)) - a(n) = 4^k. As sigma(a(n)) - a(n) = 4^k then sigma(a(n)) - a(n) = 1. As a(n) is composite this has no solutions. (End)
For the first 43 terms, a(1) is the only square term and for the other terms, a(n) has a squarefree odd part. However, this is not always true as 44088037271892 (see above) is a term and its odd part is not squarefree. - Chai Wah Wu, May 19 2025
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..43 (calculated from the b-file at A279731)
EXAMPLE
12 is not prime; 12 has proper divisors 1, 2, 3, 4, and 6, with a sum of 16. This is a square number as well as a power of 2.
MAPLE
filter:= proc(n) local s;
s:= numtheory:-sigma(n)-n;
s > 1 and s = 4^padic:-ordp(s, 4)
end proc:
select(filter, [$4..10^7]); # Robert Israel, May 13 2025
MATHEMATICA
Zweierpotenzen = {};
Quadratzahlen = {};
Beides = {};
For[k = 1, k <= 50000000, k++,
SumET = Total[Divisors[k]] - k;
If[IntegerQ[Log[2, SumET]] && PrimeQ[k] == False,
AppendTo[Zweierpotenzen, k]];
If[IntegerQ[Sqrt[SumET]] && PrimeQ[k] == False,
AppendTo[Quadratzahlen, k ]]];
Beides = Intersection[Zweierpotenzen, Quadratzahlen];
Beides
PROG
(PARI) isok(k) = if (!ispseudoprime(k), my(s=sigma(k)-k, z); issquare(s) && (ispower(s, , &z) && (z==2))); \\ Michel Marcus, May 13 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Hans Ulrich Keller, May 13 2025
EXTENSIONS
a(27)-a(29) from Michel Marcus, May 13 2025
a(30)-a(34) from Amiram Eldar, May 13 2025
STATUS
approved