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A247248
a(n) is the least k such that n divides 2^k + k.
6
1, 2, 1, 4, 4, 2, 6, 8, 7, 4, 3, 8, 12, 6, 7, 16, 16, 14, 18, 4, 19, 8, 22, 8, 33, 12, 7, 40, 11, 26, 23, 32, 8, 16, 6, 32, 5, 18, 37, 24, 40, 38, 42, 8, 7, 22, 10, 32, 61, 84, 38, 12, 35, 32, 46, 40, 32, 28, 24, 44, 17, 30, 61, 64, 66, 8, 66, 16, 67, 6, 11, 32
OFFSET
1,2
COMMENTS
For every positive integer n, there exists an integer k such that 2^k + k is divisible by n. The proof is given in the link, p. 63.
LINKS
International Mathematical Olympiad, Problem N7, IMO-2006, p. 63.
EXAMPLE
a(7) = 6 because 2^6 + 6 = 70 is divisible by 7.
MAPLE
f:= proc(n) local k;
for k from 1 do
if 2^k + k mod n = 0 then return k fi
od
end proc:
seq(f(n), n=1..100); # Robert Israel, Dec 01 2014
MATHEMATICA
Table[s=0; k=0; While[k++; s=Mod[2^k+k, n]; s>0]; k, {n, 50}]
lk[n_]:=Module[{k=1}, While[Mod[2^k+k, n]!=0, k++]; k]; Array[lk, 120] (* Harvey P. Dale, Jun 18 2022 *)
PROG
(Python)
def A247248(n):
....if n == 1:
........return 1
....else:
........x, k, kr = 1, 0, 0
........while (x+kr) % n:
............x, kr = (2*x) % n, (kr+1) % n
............k += 1
........return k
# Chai Wah Wu, Dec 03 2014
(PARI) a(n) = for(m=1, oo, if(Mod(2, n)^m==-m, return(m))); \\ Jinyuan Wang, Mar 15 2020
CROSSREFS
Cf. A135366, A006127 (2^n+n).
Sequence in context: A249140 A113421 A135366 * A192017 A180566 A051289
KEYWORD
nonn
AUTHOR
Michel Lagneau, Dec 01 2014
STATUS
approved