OFFSET
1,2
COMMENTS
For every m positive m <> 4, m^2 >= phi(m) * tau(m)^2 with equality iff m = 1, 2, 8, 12 that are the first four terms of this sequence (see De Koninck & Mercier in References).
As phi(m) is even for m >= 2, the terms > 1 of this sequence are necessarily even.
The first few quotients obtained are 1, 1, 1, 1, 2, 2, 4, 3, 9, 6, ...
The numbers of the form m = 3 * 2^(2^k-1), k >= 4 are terms. - Marius A. Burtea, Jan 17 2020
... with m^2 / (phi(m) * tau(m)^2) = 9 * 2^(2^k-2*k-3). - Bernard Schott, Jan 17 2020
REFERENCES
J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 632 pp. 83, 283, Ellipses, Paris, 2004.
EXAMPLE
96 is a term because tau(96) = 12, phi(96) = 32 and 96^2 / (32*12^2) = 2.
MAPLE
with(numtheory): filter:= m-> irem(m^2, phi(m)*tau(m)^2)=0:
select(filter, [$1..160000])[]; \\ Alois P. Heinz, Jan 17 2020
MATHEMATICA
Select[Range[160000], Divisible[#^2, EulerPhi[#] * DivisorSigma[0, #]^2] &] (* Amiram Eldar, Jan 17 2020 *)
PROG
(PARI) isok(m) = Mod(m, eulerphi(m)*numdiv(m)^2)^2 == 0; \\ Michel Marcus, Jan 17 2020
(Magma) [1] cat [m:m in [2..160000 by 2]| m^2 mod (EulerPhi(m)*DivisorSigma(0, m)^2) eq 0]; // Marius A. Burtea, Jan 17 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Jan 17 2020
STATUS
approved