%N Numbers m such that phi(m)/tau(m) is a square of an integer where phi is the Euler totient function (A000010) and tau is the number of divisors function (A000005).
%C The first 11 terms of this sequence are also the first 11 terms of A341938: m such that phi(m)*tau(m) is a square, then, a(12) = 57 while A341938(12) = 54. 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 54 (last example).
%C Some subsequences (see examples):
%C -> The seven terms that satisfy also tau(m) = phi(m) form the subsequence A020488 with phi(m)/tau(m) = 1^2.
%C -> 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 phi(p)/tau(p) = k^2.
%C -> Cubes p^3 where p is a prime of the form k^2+1 (A002496) form another subset because if p = 2, phi(8)/tau(8)=1, and if p odd, phi(p^3)/tau(p^3) = (k*p/2)^2 with k even.
%e phi(30) = 8, tau(30) = 8 so phi(30)/tau(30) = 1^2, and 30 is a term.
%e phi(45) = 24, tau(45) = 6, so phi(45)/tau(45) = 4 = 2^2, and 85 is a term.
%e phi(125) = 100, tau(125) = 4, so phi(125)/tau(125) = 25 = 5^2, and 125 is a term.
%e phi(54) = 18, tau(54) = 8, and phi(54)/tau(54) = 18/8 = 9/4 = (3/2)^2 and 54 is not a term while phi(54)*tau(54) = 12^2.
%p with(numtheory): filter:= q -> phi(q)/tau(q) = floor(phi(q)/tau(q)) and issqr(phi(q)/tau(q)) : select(filter, [$1..750]);
%t Select[Range, IntegerQ @ Sqrt[EulerPhi[#]/DivisorSigma[0, #]] &] (* _Amiram Eldar_, Feb 24 2021 *)
%o (PARI) isok(m) = my(x=eulerphi(m)/numdiv(m)); (denominator(x)==1) && issquare(x); \\ _Michel Marcus_, Feb 24 2021
%Y Intersection of A020491 and A341938.
%Y Similar for: A144695 (sigma(n)/tau(n) perfect square), A293391 (sigma(n)/phi(n) perfect square).
%Y Subsequences: A090698, A020488.
%Y Cf. A069237, A289585, A341940.
%Y Cf. A000005 (phi), A000010(tau).
%A _Bernard Schott_, Feb 24 2021