OFFSET
1,1
COMMENTS
Subsequence of A262406.
If n is a Carmichael number, then phi(n) = Product_{primes p dividing n} (p-1).
So the question is: What are the Carmichael numbers n such that Product_{primes p dividing n} (p-1) 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..10000
William D. Banks, Carmichael Numbers with a Square Totient, Canad. Math. Bull. 52 (2009), 3-8.
EXAMPLE
1729 is a term because A000010(1729) = 1729*(1-1/7)*(1-1/13)*(1-1/19) = 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
KEYWORD
nonn
AUTHOR
Altug Alkan, May 06 2016
EXTENSIONS
a(30) corrected by Amiram Eldar, Aug 11 2017
STATUS
approved