A247248 a(n) is the least k such that n divides 2^k + k.

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

Author

Michel Lagneau, Dec 01 2014

Keywords

• nonn

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.

Examples

			a(7) = 6 because 2^6 + 6 = 70 is divisible by 7.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000
• International Mathematical Olympiad, Problem N7, IMO-2006, p. 63.

Crossrefs

Cf. A135366, A006127 (2^n+n).

Programs

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 *)

PARI

a(n) = for(m=1, oo, if(Mod(2, n)^m==-m, return(m))); \\ Jinyuan Wang, Mar 15 2020

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