A383044 Numbers m such that phi(m) + phi(m+phi(m)) = m where phi is the Euler totient function.

4, 6, 8, 10, 12, 14, 16, 20, 24, 28, 32, 40, 48, 56, 64, 70, 80, 94, 96, 112, 128, 140, 160, 188, 192, 224, 256, 280, 320, 376, 384, 448, 512, 560, 640, 752, 768, 896, 1024, 1120, 1280, 1504, 1536, 1792, 2048, 2240, 2560, 3008, 3072, 3584, 4096, 4480, 5120, 6016, 6144, 7168, 8192, 8960

Offset

1,1

Comments

Empirical observation: Let phi(m) + phi(m + phi(m)) = A*m / B, GCD(A,B) = 1. For some (A,B) like (1,1) - this sequence, (2,3), (4,5), (4,7), (5,7), (7,9), (14,9), (8,11), ..., there exists (finitely/infinitely many ?) solutions to phi(m) + phi(m + phi(m)) = A*m / B. Experimentally it looks like for m = 3*A033845(n) = 18*A003586(n), phi(m) + phi(m + phi(m)) = 7*m / 9. - Ctibor O. Zizka, Apr 25 2025

Empirical observation: for a(n) <= 2^24, this sequence is the union of {2^k : k > 1}, {2^k*3 : k > 0}, {2^k*5 : k > 0}, {2^k*7 : k > 0}, {2^k*5*7 : k > 0}, {2^k*47 : k > 0}. - Michael De Vlieger, Sep 09 2025

Links

Michael De Vlieger, Table of n, a(n) for n = 1..123

Thomas Bloom, Problem 411, Erdős Problems.

Erdős problems database contributors, Erdős problem database, see no. 411.

Stefan Steinerberger, On an iterated arithmetic function problem of Erdos and Graham, arXiv:2504.08023 [math.NT], 2025.

Mathematica

q[m_] := Module[{phi = EulerPhi[m]}, phi + EulerPhi[m + phi] == m]; Select[Range[10000], q] (* Amiram Eldar, Apr 14 2025 *)

Prog

(PARI) isok(m) = eulerphi(m) + eulerphi(m+eulerphi(m)) == m;

Crossrefs

Cf. A000010, A121048.

Sequence in context: A279040 A141109 A333197 * A289426 A186331 A258036

Adjacent sequences: A383041 A383042 A383043 * A383045 A383046 A383047

Keyword

nonn

Author

Michel Marcus, Apr 14 2025

Status

approved