OFFSET
1,1
COMMENTS
The valid values of m in the equation are the terms of the sequence A151999 in order.
m is a solution if all squarefree divisors of x also divide m.
The formula is recursive. For example, taking a151999(68) we get the following: 11664=phi(108*324), 1259712=phi(11664*324), 136048896=phi(1259712*324), ...
If a solution exists then x^(k+1) = phi(x^k * m) for a fixed m, and the smallest value of k must be 1. This follows from a|b implies phi(a)|phi(b), and for k >= 1 a^(k-1)|a^k.
The smallest solution where solutions exist are the terms of the sequence A055744 not in order.
The values of phi(m) are the terms of the sequence A068997 not in order.
LINKS
Max Alekseyev, PARI scripts for various problems
FORMULA
0 < (phi(m)^(k+1) = phi(phi(m)^k*m)), k >= 1, m >= 1.
EXAMPLE
The first 1 is a term since there is only 1 solution when phi(m)=6. The solution is m=18.
The first 5 is a term since there are 5 solutions when phi(m)=16. These are 32, 34, 40, 48, and 60.
From Michel Marcus, Nov 08 2017: (Start)
Illustration of first few terms:
1: [1, 2],
2: [4, 6],
4: [8, 10, 12],
6: [18],
8: [16, 20, 24, 30],
12: [36, 42],
16: [32, 34, 40, 48, 60],
18: [54],
20: [50],
24: [72, 78, 84, 90],
32: [64, 68, 80, 96, 102, 120],
... (End)
PROG
(PARI) isok(n) = {iv = invphi(n); if (#iv, return (sum(m=1, #iv, n^2 == eulerphi(n*iv[m])))); return (0); }
lista(nn) = {for (n=1, nn, if (v = isok(n), print1(v, ", ")); ); } \\ \\ using the invphi script by Max Alekseyev; Michel Marcus, Nov 07 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Torlach Rush, Nov 05 2017
STATUS
approved