

A272798


Carmichael numbers n such that Euler totient function of n (phi(n)) is a perfect square.


4



1729, 63973, 75361, 172081, 278545, 340561, 658801, 997633, 1773289, 3224065, 5310721, 8719309, 8719921, 12945745, 13187665, 15888313, 17586361, 27402481, 29020321, 39353665, 40430401, 49333201, 67371265, 84417985, 120981601, 128697361, 129255841, 130032865, 151530401, 151813201, 158864833
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OFFSET

1,1


COMMENTS

Subsequence of A262406.
If n is a Carmichael number, then phi(n) = Product_{primes p dividing n} (p1).
So the question is: What are the Carmichael numbers n such that Product_{primes p dividing n} (p1) is a square?
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, ...
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.
This sequence is infinite. See links section for more details.  Altug Alkan, Jan 16 2017


LINKS

Amiram Eldar, Table of n, a(n) for n = 1..1000
W. D. Banks, Carmichael Numbers with a Square Totient, Canad. Math. Bull. 52(2009), 38.


EXAMPLE

1729 is a term because A000010(1729) = 1729*(11/7)*(11/13)*(11/19) = 1296 = 36^2.


PROG

(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}
lista(nn) = for(n=1, nn, if(isA002997(n) && issquare(eulerphi(n)), print1(n, ", ")));


CROSSREFS

Cf. A000010, A000290, A002997, A039770, A262406.
Sequence in context: A265328 A265628 A306479 * A212920 A317126 A318646
Adjacent sequences: A272795 A272796 A272797 * A272799 A272800 A272801


KEYWORD

nonn


AUTHOR

Altug Alkan, May 06 2016


EXTENSIONS

a(30) corrected by Amiram Eldar, Aug 11 2017


STATUS

approved



