OFFSET
1,2
COMMENTS
This is a necessary condition to have sigma(sigma(m))/sigma(m) = sigma(m)/m.
Are there other integers than 1, for which this is satisfied?
If m is an odd number such that sigma(sigma(m^2))/2 is a square, and p is in A000043 such that 2^p-1 does not divide sigma(m^2), then 2^(p-1)*m^2 is in the sequence. Such m include 5, 19, 161, 543, 1031, 1899, 3035, 6673. Thus if A000043 is infinite, so is this sequence. - Robert Israel, Aug 17 2018
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..5000 (first 200 terms from Robert Israel)
MAPLE
filter:= proc(n) local t; t:= (numtheory:-sigma @@2)(n)/n; issqr(numer(t)) and issqr(denom(t)) end proc:select(filter, [$1..200000]); # Robert Israel, Aug 17 2018
MATHEMATICA
Select[Range[10^5], IntegerQ@ Sqrt[ DivisorSigma[1, DivisorSigma[1, #]] #] &] (* Giovanni Resta, Aug 19 2018 *)
PROG
(PARI) isok(n) = issquare(sigma(sigma(n))/n);
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Aug 15 2018
STATUS
approved