%I #30 Apr 04 2019 10:30:44
%S 12,48,63,76,108,192,292,304,432,567,652,768,873,972,1168,1216,1359,
%T 1728,2107,2608,3072,3087,3532,3888,4383,4525,4612,4672,4864,5103,
%U 5409,5836,6543,6912,7204,7857,8716,8748,10372,10432,12231,12288
%N Numbers k with exactly two distinct prime factors and such that phi(k) is a square, when: k = p^(2s) * q^(2t+1) with s >= 1, t >= 0, p <> q primes.
%C An integer belongs to this sequence iff p*(p-1)*(q-1) = m^2.
%C This is the second subsequence of A324745, the first one is A324746.
%C Some values of (k,p,q,m): (12,2,3,2), (63,3,7,6), (76,2,19,6), (292,2,73,12), (652,2,163,18), (873,3,97,24).
%C The primitive terms of this sequence are the products p^2 * q, with p<>q which satisfy p*(p-1)*(q-1) = m^2. The first few primitive terms are: 12, 63, 76, 292, 652, 873.. Then the integers (p^2 * q) * p^2 and (p^2 * q) * q^2 are new terms of the general sequence.
%H David A. Corneth, <a href="/A324747/b324747.txt">Table of n, a(n) for n = 1..9878</a> (terms <= 10^10)
%F phi(p^2 * q) = p*(p-1)*(q-1) = m^2 for primitive terms.
%F phi(k) = (p^(s-1) * q^t * m)^2 with k as in the name of this sequence.
%e 63 = 3^2 * 7 and phi(63) = 3*2*6 = 6^2.
%e 1728 = 2^6 * 3^3 and phi(1728) = (2^2 * 3^1 * 2)^2 = 24^2.
%o (PARI) isok(k) = {if (issquare(eulerphi(k)), my(expo = factor(k)[,2]); if ((#expo == 2)&& (expo[1]%2) != (expo[2]%2), return (1)););} \\ _Michel Marcus_, Mar 18 2019
%Y Cf. A039770, A054755, A324745, A324746.
%K nonn
%O 1,1
%A _Bernard Schott_, Mar 13 2019
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