

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

This is the second subsequence of A323916, the first one is A323917.
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

Table of n, a(n) for n=1..47.


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

Cf. A000396, A051953, A063752, A246551, A323916, A323917, A306670.
Sequence in context: A290543 A183230 A246911 * A039460 A245990 A110829
Adjacent sequences: A323915 A323916 A323917 * A323919 A323920 A323921


KEYWORD

nonn


AUTHOR

Bernard Schott, Feb 09 2019


STATUS

approved



