A159836 Integers n such that the orbit n, f(n), f(f(n)), ... is eventually periodic with period 2, where f(n) = product(a(k)^p(k)) when n has the prime factorization n = product(p(k)^a(k)).

8, 9, 18, 24, 25, 32, 36, 40, 45, 49, 50, 56, 63, 64, 75, 81, 88, 90, 96, 98, 99, 100, 104, 117, 120, 121, 125, 126, 128, 136, 144, 147, 150, 152, 153, 160, 162, 168, 169, 171, 175, 180, 184, 192, 196, 198, 200, 207, 216, 224, 225, 232, 234, 242, 243, 245, 248

Offset: 1

Author

John W. Layman, Apr 23 2009

Keywords

• nonn

Comments

It is proved in the reference that for every positive integer n the orbit n, f(n), f(f(n)), ... is eventually periodic with period 1 or 2.

Includes all numbers whose prime exponents are distinct primes. If n is in this sequence and k is a squarefree number such that (k,n) = 1, then k*n is in this sequence. - Charlie Neder, May 16 2019

Links

• Paolo P. Lava, Table of n, a(n) for n = 1..5000
• M. Farrokhi, The Prime Exponentiation of an Integer: Problem 11315, Amer. Math. Monthly, 116 (2009), 470.

Crossrefs

Cf. A008477, A008478.

Programs

Maple

with(numtheory); P:=proc(q) local a0f,a1,a1f,a2,a2f,a3,a3f,a4,a4f,k,n;
for n from 1 to q do a0:=1;a1:=1;a2:=2;a3:=3;a4:=n;
  while not (a1=a3 and a2=a4) do a0f:=ifactors(a4)[2];
   a1:=mul(a0f[k][2]^a0f[k][1],k=1..nops(a0f)); a1f:=ifactors(a1)[2];
   a2:=mul(a1f[k][2]^a1f[k][1],k=1..nops(a1f)); a2f:=ifactors(a2)[2];
   a3:=mul(a2f[k][2]^a2f[k][1],k=1..nops(a2f)); a3f:=ifactors(a3)[2];
   a4:=mul(a3f[k][2]^a3f[k][1],k=1..nops(a3f)); od;
if a1<>a2 then print(n); fi; od; end: P(10^6); # Paolo P. Lava, Oct 24 2013

Mathematica

f[n_] := Module[{f = Transpose[FactorInteger[n]]}, Times @@ (f[[2]]^f[[1]])]; Select[Range[300], (x = NestWhileList[f, #, UnsameQ, All]; x[[-2]] != x[[-1]]) &] (* T. D. Noe, Oct 24 2013 *)