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A297475
Numbers n such that phi(x) = n for more than one value of x, and the smallest such x divides the largest.
2
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
1,2
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.
FORMULA
2 = max({phi^-1(n)}) / min({phi^-1(n)}).
0 = A006511(n) mod A002181(n).
EXAMPLE
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.
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 *)
PROG
(PARI) isok(n) = my(vx = invphi(n)); (#vx > 1) && ((vecmax(vx) % vecmin(vx)) == 0); \\ Michel Marcus, Jul 18 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Torlach Rush, Dec 30 2017
STATUS
approved