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

1,2

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

Links

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

Formula

a(n) <= A098174(n).

Example

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.

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 *)

Prog

(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

Crossrefs

Cf. A098174, A255430, A255357, A275533.

Sequence in context: A031351 A068923 A222295 * A103714 A260486 A193514

Adjacent sequences: A371703 A371704 A371705 * A371707 A371708 A371709

Keyword

nonn,base

Author

Robert Israel, Apr 03 2024

Status

approved