A331459 Integers m such that phi(m) * tau(m)^2 divides m^2.

1, 2, 8, 12, 80, 96, 128, 720, 972, 1152, 1200, 1344, 2560, 4032, 6144, 6912, 7680, 7776, 8100, 10000, 23040, 26244, 30000, 30720, 32768, 34560, 38400, 55296, 56320, 62208, 64000, 64800, 80000, 84672, 90000, 97200, 98304, 103680, 108864, 110000, 142884, 159744

Offset: 1

Author

Bernard Schott, Jan 17 2020

Keywords

• nonn

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

Examples

			96 is a term because tau(96) = 12, phi(96) = 32 and 96^2 / (32*12^2) = 2.
		

References

• J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 632 pp. 83, 283, Ellipses, Paris, 2004.

Crossrefs

Cf. A000005 (tau), A000010 (phi), A022157.

Programs

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

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 *)

PARI

isok(m) = Mod(m, eulerphi(m)*numdiv(m)^2)^2 == 0; \\ Michel Marcus, Jan 17 2020