OFFSET
1,1
COMMENTS
All numbers 3*2^k k>=2 are in the sequence.
Let n=p^k*q where p^k is the largest prime power is the factorization of n and (p,q)=1. If n belongs to the sequence then p^k = phi(n) = (p-1)*p^(k-1)*phi(q), implying that p=2 (since p-1 cannot divide p^k for prime p>2). Then 2 = phi(q), implying that q=3. Therefore the terms are simply the sequence 3*2^n for n=2,3,... - Max Alekseyev, Mar 02 2007
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).
FORMULA
a(n) = 3*2^(n+1).
MATHEMATICA
Table[3*2^(n + 1), {n, 1, 30}] (* Stefan Steinerberger, Jun 17 2007 *)
NestList[2#&, 12, 30] (* Harvey P. Dale, Jun 05 2024 *)
PROG
(Magma) [3*2^(n + 1): n in [1..35]]; // Vincenzo Librandi, May 18 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Benoit Cloitre, Apr 10 2003
STATUS
approved