A217157 a(n) is the least value of k such that the decimal expansion of n^k contains two consecutive identical digits.

16, 11, 8, 11, 5, 6, 6, 6, 2, 1, 2, 9, 3, 2, 4, 7, 5, 5, 2, 2, 1, 6, 4, 6, 5, 4, 8, 5, 2, 6, 5, 1, 2, 2, 3, 7, 2, 4, 2, 5, 3, 4, 1, 3, 2, 2, 3, 3, 2, 7, 4, 3, 6, 1, 4, 4, 2, 4, 2, 3, 2, 3, 3, 2, 1, 2, 3, 4, 2, 3, 7, 6, 3, 6, 2, 1, 3, 4, 2, 3, 3, 2, 5, 2, 4, 6

Offset: 2

Author

V. Raman, Sep 27 2012

Keywords

• nonn
• base

Comments

Least number m such that n^m is a term of A171901 - Chai Wah Wu, Feb 20 2019

Conjecture: 1 <= a(n) <= 16 for n > 1 and a(n) < 16 for n > 2. - Chai Wah Wu, Feb 20 2019

a(n) >= 1 for all n > 1 and is bounded: see link for proof. - Robert Israel, Feb 21 2019

Links

• V. Raman, Table of n, a(n) for n = 2..10000
• Robert Israel, Proof that A217157 >= 1 and is bounded

Crossrefs

Cf. A045875, A215236.

Cf. A215727, A215728, A215729, A215730, A215731, A171901, A306305.

Programs

Maple

f:= proc(n) local L,k;
   for k from 1 do
      L:= convert(n^k,base,10);
      if has(L[2..-1]-L[1..-2],0) then return k fi
   od
end proc:
map(f, [$2..100]); # Robert Israel, Feb 21 2019

Mathematica

Table[k = 1; While[! MemberQ[Differences[IntegerDigits[n^k]], 0], k++]; k, {n, 2, 100}] (* T. D. Noe, Oct 01 2012 *)

Python

def A217157(n):
    m, k = 1, n
    while True:
        s = str(k)
        for i in range(1,len(s)):
            if s[i] == s[i-1]:
                return m
        m += 1
        k *= n # Chai Wah Wu, Feb 20 2019

Formula

a(A171901(n)) = 1. - Chai Wah Wu, Feb 20 2019

a(n) = A215236(n) + 1. - Georg Fischer, Nov 25 2020