A371706 a(n) is the least k > 0 such that n^k contains the digit 1.

1, 4, 4, 2, 3, 3, 4, 3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 3, 3, 3, 3, 3, 3, 3, 2, 4, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 1, 3, 3, 2, 3, 2, 3, 3, 2, 3, 1, 4, 4, 3, 4, 4, 4, 3, 2, 4, 1, 2, 3, 5, 3, 4, 4, 4, 2, 3, 1, 3, 3, 4, 3, 4, 4, 3, 2, 2, 1, 4, 4, 6, 4, 2, 3, 3, 2

Offset: 1

Author

Robert Israel, Apr 03 2024

Keywords

• nonn
• base

Comments

First n such that a(n) = k: 1, 4, 5, 2, 74, 94, 305, 2975 for k = 1 to 8.

a(n) <= 8 for n <= 240000000. Is a(n) ever greater than 8?

a(n) <= 8 for n <= 10^11. Moreover the only n <= 10^11 with a(n) = 8 are 2975 * 10^j. - Robert Israel, Apr 05 2024

a(n) <= 8 for n <= 10^13, and a(n) <= 17 for all n; see A275533. - Michael S. Branicky, Apr 08 2024

Examples

			a(3) = 4 because 3^1, 3^2 = 9 and 3^3 = 27 have no 1's, but 3^4 = 81 does have a 1.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A098174, A255430, A255357, A275533.

Programs

Maple

g:= proc(n) local k;
   for k from 1 do if member(1,convert(n^k,base,10)) then return k fi od;
end proc:
map(g, [$1..100]);

Mathematica

seq={};Do[k=1;While[!ContainsAny[IntegerDigits[n^k],{1}],k++];AppendTo[seq,k],{n,99}];seq (* James C. McMahon, Apr 05 2024 *)

Python

from itertools import count
def A371706(n):
    m = n
    for k in count(1):
        if '1' in str(m):
            return k
        m *= n # Chai Wah Wu, Apr 04 2024

Formula

a(n) <= A098174(n).