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 A081808 Numbers n such that the largest prime power in the factorization of n equals phi(n). 2
 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368, 1610612736, 3221225472 (list; graph; refs; listen; history; text; internal format)
 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 *) PROG (MAGMA) [3*2^(n + 1): n in [1..35]]; // Vincenzo Librandi, May 18 2011 CROSSREFS Essentially the same as A007283 = 3*2^n. Sequence in context: A181924 A270257 A180617 * A260261 A080495 A090776 Adjacent sequences:  A081805 A081806 A081807 * A081809 A081810 A081811 KEYWORD nonn,easy AUTHOR Benoit Cloitre, Apr 10 2003 STATUS approved

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Last modified October 20 12:47 EDT 2019. Contains 328257 sequences. (Running on oeis4.)