A341939 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).

1, 3, 8, 10, 18, 19, 24, 30, 34, 45, 52, 57, 73, 74, 85, 102, 125, 135, 140, 152, 153, 156, 163, 182, 185, 190, 202, 219, 222, 252, 255, 333, 342, 360, 375, 394, 416, 420, 436, 451, 455, 456, 459, 476, 489, 505, 514, 546, 555, 570, 584, 606, 625, 629, 640, 646, 679, 680, 730

Offset: 1

Author

Bernard Schott, Feb 24 2021

Keywords

• nonn

Comments

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).

Some subsequences (see examples):

-> The seven terms that satisfy also tau(m) = phi(m) form the subsequence A020488 with phi(m)/tau(m) = 1^2.

-> 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.

-> 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.

Examples

			phi(30) = 8, tau(30) = 8 so phi(30)/tau(30) = 1^2, and 30 is a term.
phi(45) = 24, tau(45) = 6, so phi(45)/tau(45) = 4 = 2^2, and 85 is a term.
phi(125) = 100, tau(125) = 4, so phi(125)/tau(125) = 25 = 5^2, and 125 is a term.
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.
		

Crossrefs

Intersection of A020491 and A341938.

Similar for: A144695 (sigma(n)/tau(n) perfect square), A293391 (sigma(n)/phi(n) perfect square).

Subsequences: A090698, A020488.

Cf. A069237, A289585, A341940.

Cf. A000005 (phi), A000010(tau).

Programs

Maple

with(numtheory): filter:= q -> phi(q)/tau(q) = floor(phi(q)/tau(q)) and issqr(phi(q)/tau(q)) : select(filter, [$1..750]);

Mathematica

Select[Range[1000], IntegerQ @ Sqrt[EulerPhi[#]/DivisorSigma[0, #]] &] (* Amiram Eldar, Feb 24 2021 *)

PARI

isok(m) = my(x=eulerphi(m)/numdiv(m)); (denominator(x)==1) && issquare(x); \\ Michel Marcus, Feb 24 2021