A297475 Numbers n such that phi(x) = n for more than one value of x, and the smallest such x divides the largest.

1, 2, 8, 10, 22, 28, 30, 44, 46, 52, 54, 56, 58, 66, 70, 78, 82, 92, 102, 104, 106, 110, 116, 126, 128, 130, 136, 138, 140, 148, 150, 164, 166, 172, 178, 190, 196, 198, 204, 210, 212, 222, 226, 228, 238, 250, 260, 262, 268, 270, 282, 292, 294, 296, 306, 310, 316, 330, 332, 342, 344, 346, 356, 358, 366, 368, 372

Offset: 1

Author

Torlach Rush, Dec 30 2017

Keywords

• nonn

Comments

The larger endpoint is always twice the value of the smaller endpoint.

Conjecture 1: The number of solutions, excluding endpoints is always 0, or an odd number. (known to n = 2 * 10^5)

Conjecture 2: If both endpoints are divisible by 5, then the number of solutions (excluding terms of A007366) is of the form 4k + 1. (known to n = 2 * 10^5)

A007366 is contained in this sequence and the number of solutions, excluding endpoints is always 0.

Terms of this sequence are totients with a single odd totient inverse.

Examples

			2 is in the sequence because {phi^-1(2)} = {3,4,6}, and 2 = 6 / 3.
8 is in the sequence because {phi^-1(8)} = {15,...,30}, and 2 = 30 / 15.
10 is in the sequence because {phi^-1(10)} = {11,22}, and 2 = 22 / 11.
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..2441
• Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Crossrefs

Cf. A002181, A002202, A006511, A007366, A032447.

Programs

Mathematica

With[{nn = 67}, Take[#, nn] &@ Keys@ Select[KeySort@ PositionIndex@ Array[EulerPhi, nn^2], IntegerQ[#2/#1] & @@ {First@ #, Last@ #} &]] (* Michael De Vlieger, Dec 31 2017 *)

PARI

isok(n) = my(vx = invphi(n)); (#vx > 1) && ((vecmax(vx) % vecmin(vx)) == 0); \\ Michel Marcus, Jul 18 2018

Formula

2 = max({phi^-1(n)}) / min({phi^-1(n)}).

0 = A006511(n) mod A002181(n).