A323917 Numbers k with exactly two distinct prime divisors and such that cototient(k) is square, where k = p^(2s+1) * q^(2t+1) with s,t >=0, p, q primes and p + q - 1 = M^2.

6, 21, 24, 54, 69, 96, 133, 141, 189, 216, 237, 301, 384, 481, 486, 501, 589, 621, 669, 781, 864, 1029, 1077, 1141, 1269, 1317, 1357, 1417, 1536, 1537, 1701, 1944, 1957, 1981, 2041, 2133, 2181, 2517, 2869, 3261, 3397, 3456, 3601, 3661, 3669, 4101, 4309, 4333, 4374, 4509

Offset: 1

Author

Bernard Schott, Feb 09 2019

Keywords

• nonn

Comments

This is the first subsequence of A323916, the second one is A323918 and A323916 = {this sequence} Union A323918 with empty intersection.

Some values of (k,p,q,M): (6,2,3,2), (21,3,7,3), (69,3,23,5), (133,7,19,5), (141,3,47,9), (301,7,43,7), (481,13,37,7).

The primitive terms of this sequence are the products p * q, with p,q which satisfy p+q-1 = M^2, the first ones are: 6, 21, 69, 133, 141, 237. Then the integers (p*q) * p^2 and (p*q) * q^2 are new terms of the general sequence.

There is only one even perfect number in this sequence: 6. The other ones are in A323918.

See the file "Subsequences and Subfamilies of terms" (&2.1) in A063752 for more details, proofs with data, comments, formulas and examples.

Examples

			Perfect number 6 = 2 * 3 and cototient(6) = 2^2.
781 = 11 * 71 and cototient(781) = 11 + 71 - 1 = 9^2.
864 = 2^5 * 3^3 and cototient(864)= (2^2 * 3^1 * 2)^2 = 24^2.
		

Crossrefs

Cf. A051953, A063752, A246551, A323916, A323918.

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*q) = p + q - 1 = M^2 for primitive terms.

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