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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.
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
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+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.
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.
EXAMPLE
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.
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
KEYWORD
nonn
AUTHOR
Bernard Schott, Feb 09 2019
STATUS
approved