

A349767


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


3



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
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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 19691993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1983, page 158, 1993.


LINKS

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


MAPLE

filter:= n > 2^nn 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


CROSSREFS

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


KEYWORD

nonn


AUTHOR



STATUS

approved



