|
|
A293928
|
|
Totients phi(m) having one or more solutions to phi(m)^(k+1) = phi(phi(m)^k*m), k >= 1, m >= 1.
|
|
2
|
|
|
1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 32, 36, 40, 48, 54, 64, 72, 80, 84, 96, 100, 108, 120, 128, 144, 160, 162, 168, 192, 200, 216, 240, 252, 256, 272, 288, 312, 320, 324, 336, 360, 384, 400, 432, 440, 480, 486, 500, 504, 512, 544, 576, 588, 600, 624, 640, 648, 672, 684
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
The smallest totient absent from the list is 10. This is because the totient inverses of 10, 11 and 22 are not solutions of phi(m)^(k+1) = phi(phi(m)^k*m), k >= 1, m >= 1.
The formula is recursive. For example, taking a(22) we get the following: 11664 = phi(108*324), 1259712 = phi(11664*324), 136048896 = phi(1259712*324), ...
If a solution exists then 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.
|
|
LINKS
|
|
|
FORMULA
|
0 < phi(m)^(k+1) = phi(phi(m)^k*m), k >= 1, m >= 1.
|
|
EXAMPLE
|
96 is a term since 96^2 = phi(96*288), with k=1 and m=288 where phi(288) = 96.
|
|
PROG
|
(PARI) isok(n) = {my(iv = invphi(n)); if (#iv, for (m = 1, #iv, if (n^2 == eulerphi(n*iv[m]), return (1)); ); ); return (0); } \\ using the invphi script by Max Alekseyev; Michel Marcus, Nov 01 2017
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|