|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
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
|
|
|
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
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|