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

Author

Michel Marcus, Apr 14 2025

Keywords

• nonn

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

Links

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

Crossrefs

Cf. A000010, A121048.

Programs

Mathematica

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

PARI

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