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+q-1) = M^2.

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

Offset: 1

Author

Bernard Schott, Feb 09 2019

Keywords

• nonn

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+q-1) = 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.

Examples

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

Crossrefs

Cf. A000396, A051953, A063752, A246551, A323916, A323917, A306670.

Programs

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

Formula

cototient(p^2 * q) = p * (p + q - 1) = M^2;

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