A349767 Numbers m such that 2^m - m is divisible by 5.

3, 14, 16, 17, 23, 34, 36, 37, 43, 54, 56, 57, 63, 74, 76, 77, 83, 94, 96, 97, 103, 114, 116, 117, 123, 134, 136, 137, 143, 154, 156, 157, 163, 174, 176, 177, 183, 194, 196, 197, 203, 214, 216, 217, 223, 234, 236, 237, 243, 254, 256, 257, 263, 274, 276, 277, 283, 294, 296, 297, 303

Offset: 1

Author

Bernard Schott, Dec 10 2021

Keywords

• nonn

Comments

For every prime p, there are infinitely many numbers m such that 2^m - m (A000325) is divisible by p, here are numbers m corresponding to p = 5.

Equivalently, numbers that are congruent to {3, 14, 16, 17, 23, 34, 36, 37, 43, 54, 56, 57} mod 60, <==> numbers that are congruent to {+-3, +-14, +-16, +-17, +-23, +-34} mod 60.

References

• Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1983, page 158, 1993.

Links

• The IMO Compendium, Problem 4, 15th Canadian Mathematical Olympiad 1983.
• Index to sequences related to Olympiads.

Crossrefs

Cf. A000325, A070402, A010874.

Similar with: A299174 (p = 2), A047257 (p = 3), this sequence (p = 5).

Programs

Maple

filter:= n -> 2^n-n mod 5 = 0 : select(filter, [$1..400]);

Mathematica

Select[Range[300], PowerMod[2, #, 5] == Mod[#, 5] &] (* Amiram Eldar, Dec 10 2021 *)

PARI

isok(m) = Mod(2, 5)^m == Mod(m, 5); \\ Michel Marcus, Dec 10 2021

Python

def ok(n): return pow(2, n, 5) == n%5
print([k for k in range(357) if ok(k)]) # Michael S. Branicky, Dec 10 2021