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%I #34 Aug 18 2017 18:18:50
%S 1729,63973,75361,172081,278545,340561,658801,997633,1773289,3224065,
%T 5310721,8719309,8719921,12945745,13187665,15888313,17586361,27402481,
%U 29020321,39353665,40430401,49333201,67371265,84417985,120981601,128697361,129255841,130032865,151530401,151813201,158864833
%N Carmichael numbers n such that Euler totient function of n (phi(n)) is a perfect square.
%C Subsequence of A262406.
%C If n is a Carmichael number, then phi(n) = Product_{primes p dividing n} (p-1).
%C So the question is: What are the Carmichael numbers n such that Product_{primes p dividing n} (p-1) is a square?
%C The number of prime divisors of terms of this sequence are 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 5, 5, 5, 4, 4, 4, 4, 4, ...
%C 1299963601 = 601*1201*1801 is the second term that has three prime divisors and it is a member of this sequence since 600*1200*1800 = 2^10*3^4*5^6 is a square.
%C This sequence is infinite. See links section for more details. - _Altug Alkan_, Jan 16 2017
%H Amiram Eldar, <a href="/A272798/b272798.txt">Table of n, a(n) for n = 1..1000</a>
%H W. D. Banks, <a href="http://dx.doi.org/10.4153/CMB-2009-001-7">Carmichael Numbers with a Square Totient</a>, Canad. Math. Bull. 52(2009), 3-8.
%e 1729 is a term because A000010(1729) = 1729*(1-1/7)*(1-1/13)*(1-1/19) = 1296 = 36^2.
%o (PARI) isA002997(n) = {my(f); bittest(n, 0) && !for(i=1, #f=factor(n)~, (f[2, i]==1 && n%(f[1, i]-1)==1)||return) && #f>1}
%o lista(nn) = for(n=1, nn, if(isA002997(n) && issquare(eulerphi(n)), print1(n, ", ")));
%Y Cf. A000010, A000290, A002997, A039770, A262406.
%K nonn
%O 1,1
%A _Altug Alkan_, May 06 2016
%E a(30) corrected by _Amiram Eldar_, Aug 11 2017
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