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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
OFFSET
1,2
COMMENTS
"Totients" are terms of A000010. - N. J. A. Sloane, Oct 22 2017
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.
Where (if ever) does this first differ from A068997? - R. J. Mathar, Oct 30 2017
Apparently the set of the m is A151999. - R. J. Mathar, Mar 25 2024
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
Subsequence of A002202.
Sequence in context: A325763 A363949 A068997 * A067712 A060765 A140110
KEYWORD
nonn
AUTHOR
Torlach Rush, Oct 19 2017
EXTENSIONS
More terms from Michel Marcus, Oct 24 2017
STATUS
approved