OFFSET
1,2
COMMENTS
The first 11 terms of this sequence are also the first 11 terms of A341939: m such that phi(m)/tau(m) is the square of an integer. Indeed, if phi(m)/tau(m) is a perfect square then phi(m)*tau(m) is also a square, but the converse is false. These counterexamples are in A341940, the first one is a(12) = 54.
If k and q are terms and coprimes, then k*q is another term.
Some subsequences (see examples):
-> The seven terms that satisfy tau(m) = phi(m) form the subsequence A020488.
-> Primes p of the form 2*k^2 + 1 (A090698) form another subsequence because tau(p) = 2 and phi(p) = p-1 = 2*k^2, so tau(p)*phi(p) = (2*k)^2.
-> Cubes p^3 where p is a prime of the form k^2+1 (A002496) form another subset with tau(p^3)*phi(p^3) = (2*k*p)^2.
EXAMPLE
phi(18) = tau(18) = 6, so phi(18)*tau(18) = 6^2.
phi(19) = 18, tau(19) = 2, so phi(19)*tau(19) = 36 = 6^2.
phi(34) = 16, tau(34) = 4, so phi(34)*tau(34) = 16*4 = 64 = 8^2.
phi(125) = 100, tau(125) = 4, so phi(125)*tau(125) = 400 = 20^2.
MAPLE
with(numtheory): filter:= n -> issqr(phi(n)*tau(n)) : select(filter, [$1..750]);
MATHEMATICA
Select[Range[1000], IntegerQ @ GeometricMean[{DivisorSigma[0, #], EulerPhi[#]}] &] (* Amiram Eldar, Feb 24 2021 *)
PROG
(PARI) isok(m) = issquare(numdiv(m)*eulerphi(m)); \\ Michel Marcus, Feb 24 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Feb 24 2021
STATUS
approved