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

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, 6442450944, 12884901888, 25769803776

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).

From Elmo R. Oliveira, Nov 24 2025: (Start)

G.f.: 12*x/(1-2*x).

E.g.f.: 6*(exp(2*x) - 1).

a(n) = 2*a(n-1) for n > 1.

a(n) = 6*A000079(n) = 4*A007283(n-1) = 3*A020707(n-1). (End)

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

Essentially the same as A007283 (3*2^n).

Cf. A000079, A020707.

Sequence in context: A270257 A367361 A180617 * A369798 A260261 A080495

Adjacent sequences: A081805 A081806 A081807 * A081809 A081810 A081811

Keyword

nonn,easy

Author

Benoit Cloitre, Apr 10 2003

Status

approved