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

1,1

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

Table of n, a(n) for n=1..61.

The IMO Compendium, Problem 4, 15th Canadian Mathematical Olympiad 1983.

Index to sequences related to Olympiads.

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 *)

Prog

(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

Crossrefs

Cf. A000325, A070402, A010874.

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

Sequence in context: A016062 A009401 A195864 * A022890 A032920 A134767

Adjacent sequences: A349764 A349765 A349766 * A349768 A349769 A349770

Keyword

nonn

Author

Bernard Schott, Dec 10 2021

Status

approved