OFFSET
1,1
COMMENTS
If k^2 + 1 is prime, trivially phi(k^2 + 1) == 0 (mod k).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..500 (terms 1..300 from Donovan Johnson)
MATHEMATICA
n2Q[n_]:=Module[{c=n^2+1}, CompositeQ[c]&&Divisible[EulerPhi[c], n]]; Select[Range[50000], n2Q] (* Harvey P. Dale, Apr 16 2015 *)
PROG
(PARI) for(n=1, 46656, m=n^2+1; if(isprime(m)==0, if(eulerphi(m)%n==0, print1(n ", ")))) \\ Donovan Johnson, Nov 14 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Feb 22 2002
STATUS
approved