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%I #29 Sep 08 2022 08:45:09
%S 12,24,48,96,192,384,768,1536,3072,6144,12288,24576,49152,98304,
%T 196608,393216,786432,1572864,3145728,6291456,12582912,25165824,
%U 50331648,100663296,201326592,402653184,805306368,1610612736,3221225472
%N Numbers n such that the largest prime power in the factorization of n equals phi(n).
%C All numbers 3*2^k k>=2 are in the sequence.
%C 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 Vincenzo Librandi, <a href="/A081808/b081808.txt">Table of n, a(n) for n = 1..1000</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (2).
%F a(n) = 3*2^(n+1).
%t Table[3*2^(n + 1), {n, 1, 30}] (* _Stefan Steinerberger_, Jun 17 2007 *)
%o (Magma) [3*2^(n + 1): n in [1..35]]; // _Vincenzo Librandi_, May 18 2011
%Y Essentially the same as A007283 = 3*2^n.
%K nonn,easy
%O 1,1
%A _Benoit Cloitre_, Apr 10 2003
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