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

%I A331459 #43 Sep 08 2022 08:46:25
%S A331459 1,2,8,12,80,96,128,720,972,1152,1200,1344,2560,4032,6144,6912,7680,
%T A331459 7776,8100,10000,23040,26244,30000,30720,32768,34560,38400,55296,
%U A331459 56320,62208,64000,64800,80000,84672,90000,97200,98304,103680,108864,110000,142884,159744
%N A331459 Integers m such that phi(m) * tau(m)^2 divides m^2.
%C A331459 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).
%C A331459 As phi(m) is even for m >= 2, the terms > 1 of this sequence are necessarily even.
%C A331459 The first few quotients obtained are 1, 1, 1, 1, 2, 2, 4, 3, 9, 6, ...
%C A331459 The numbers of the form m = 3 * 2^(2^k-1), k >= 4 are terms. - _Marius A. Burtea_, Jan 17 2020
%C A331459 ... with m^2 / (phi(m) * tau(m)^2) = 9 * 2^(2^k-2*k-3). - _Bernard Schott_, Jan 17 2020
%D A331459 J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 632 pp. 83, 283, Ellipses, Paris, 2004.
%e A331459 96 is a term because tau(96) = 12, phi(96) = 32 and 96^2 / (32*12^2) = 2.
%p A331459 with(numtheory): filter:= m-> irem(m^2, phi(m)*tau(m)^2)=0:
%p A331459 select(filter, [$1..160000])[]; \\ _Alois P. Heinz_, Jan 17 2020
%t A331459 Select[Range[160000], Divisible[#^2, EulerPhi[#] * DivisorSigma[0, #]^2] &] (* _Amiram Eldar_, Jan 17 2020 *)
%o A331459 (PARI) isok(m) = Mod(m, eulerphi(m)*numdiv(m)^2)^2 == 0; \\ _Michel Marcus_, Jan 17 2020
%o A331459 (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
%Y A331459 Cf. A000005 (tau), A000010 (phi), A022157.
%K A331459 nonn
%O A331459 1,2
%A A331459 _Bernard Schott_, Jan 17 2020