A324747 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.

12, 48, 63, 76, 108, 192, 292, 304, 432, 567, 652, 768, 873, 972, 1168, 1216, 1359, 1728, 2107, 2608, 3072, 3087, 3532, 3888, 4383, 4525, 4612, 4672, 4864, 5103, 5409, 5836, 6543, 6912, 7204, 7857, 8716, 8748, 10372, 10432, 12231, 12288

Offset

1,1

Comments

An integer belongs to this sequence iff p*(p-1)*(q-1) = m^2.

This is the second subsequence of A324745, the first one is A324746.

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).

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.

Links

David A. Corneth, Table of n, a(n) for n = 1..9878 (terms <= 10^10)

Formula

phi(p^2 * q) = p*(p-1)*(q-1) = m^2 for primitive terms.

phi(k) = (p^(s-1) * q^t * m)^2 with k as in the name of this sequence.

Example

63 = 3^2 * 7 and phi(63) = 3*2*6 = 6^2.
1728 = 2^6 * 3^3 and phi(1728) = (2^2 * 3^1 * 2)^2 = 24^2.

Prog

(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

Crossrefs

Cf. A039770, A054755, A324745, A324746.

Sequence in context: A230919 A181925 A118903 * A044114 A323008 A371599

Adjacent sequences: A324744 A324745 A324746 * A324748 A324749 A324750

Keyword

nonn

Author

Bernard Schott, Mar 13 2019

Status

approved