

A323918


Numbers k with exactly two distinct prime divisors and such that cototient(k) is a square, where: k = p^(2s) * q^(2t+1) with s >= 1, t >= 0, p <> q primes and such that p * (p+q1) = M^2.


4



28, 68, 112, 124, 272, 284, 388, 448, 496, 508, 657, 796, 964, 1025, 1088, 1136, 1348, 1372, 1552, 1792, 1796, 1984, 2032, 2169, 2308, 2588, 3184, 3524, 3856, 3868, 4352, 4544, 4604, 4996, 5392, 5488, 5913, 6025, 6057, 6208, 6268, 7168, 7184, 7936, 8128, 9232, 9244
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OFFSET

1,1


COMMENTS

Some values of (k,p,q,M): (28,2,7,2), (68,2,17,3), (124,2,31,4), (284,2,71,6), (388,97,7), (657,3,73,5).
The primitive terms of this sequence are the products p^2 * q, with p,q which satisfy p*(p+q1) = M^2; the first ones are 28, 68, 124, 284, 388, 508, 657, 796. Then, the integers (p^2 * q) * p^2 and (p^2 * q) * q^2 are new terms of the general sequence.
Except 6, all the even perfect numbers of A000396 belong to this sequence.
See the file "Subfamilies of terms" in A063752 for more details, proofs with data, comments, formulas and examples.


LINKS



FORMULA

cototient(p^2 * q) = p * (p + q  1) = M^2;
cototient(k) = (p^(s1) * q^t * M)^2 with k as in the name of this sequence.


EXAMPLE

272 = 2^4 * 17, M = 2*(2+171) = 6^2 and cototient(272) = (2^1 * 17^0 * 6)^2 = 12^2.
1025 = 5^2 * 41 and cototient(1025) = 5 * (5+411) = 15^2.
Perfect number: 8128 = 2^6 * 127 and cototient(8128) = 64^2.


PROG

(PARI) isok(n) = (omega(n)==2) && issquare(n  eulerphi(n)) && ((factor(n)[1, 2] % 2) != (factor(n)[2, 2] % 2)); \\ Michel Marcus, Feb 10 2019


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



