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%I #26 Mar 05 2019 16:29:26
%S 6,21,24,54,69,96,133,141,189,216,237,301,384,481,486,501,589,621,669,
%T 781,864,1029,1077,1141,1269,1317,1357,1417,1536,1537,1701,1944,1957,
%U 1981,2041,2133,2181,2517,2869,3261,3397,3456,3601,3661,3669,4101,4309,4333,4374,4509
%N 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.
%C This is the first subsequence of A323916, the second one is A323918 and A323916 = {this sequence} Union A323918 with empty intersection.
%C 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).
%C 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.
%C There is only one even perfect number in this sequence: 6. The other ones are in A323918.
%C See the file "Subsequences and Subfamilies of terms" (&2.1) in A063752 for more details, proofs with data, comments, formulas and examples.
%F cototient(p*q) = p + q - 1 = M^2 for primitive terms.
%F cototient(k) = (p^s * q^t * M)^2 with k as in the name of this sequence.
%e Perfect number 6 = 2 * 3 and cototient(6) = 2^2.
%e 781 = 11 * 71 and cototient(781) = 11 + 71 - 1 = 9^2.
%e 864 = 2^5 * 3^3 and cototient(864)= (2^2 * 3^1 * 2)^2 = 24^2.
%o (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
%Y Cf. A051953, A063752, A246551, A323916, A323918.
%K nonn
%O 1,1
%A _Bernard Schott_, Feb 09 2019
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