cp's OEIS Frontend

A081808 Numbers n such that the largest prime power in the factorization of n equals phi(n).

%I A081808 #31 Jun 05 2024 17:35:19
%S A081808 12,24,48,96,192,384,768,1536,3072,6144,12288,24576,49152,98304,
%T A081808 196608,393216,786432,1572864,3145728,6291456,12582912,25165824,
%U A081808 50331648,100663296,201326592,402653184,805306368,1610612736,3221225472
%N A081808 Numbers n such that the largest prime power in the factorization of n equals phi(n).
%C A081808 All numbers 3*2^k k>=2 are in the sequence.
%C A081808 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
%H A081808 Vincenzo Librandi, <a href="/A081808/b081808.txt">Table of n, a(n) for n = 1..1000</a>
%H A081808 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A081808 <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (2).
%F A081808 a(n) = 3*2^(n+1).
%t A081808 Table[3*2^(n + 1), {n, 1, 30}] (* _Stefan Steinerberger_, Jun 17 2007 *)
%t A081808 NestList[2#&,12,30] (* _Harvey P. Dale_, Jun 05 2024 *)
%o A081808 (Magma) [3*2^(n + 1): n in [1..35]]; // _Vincenzo Librandi_, May 18 2011
%Y A081808 Essentially the same as A007283 = 3*2^n.
%K A081808 nonn,easy
%O A081808 1,1
%A A081808 _Benoit Cloitre_, Apr 10 2003