A306305 -id:A306305 - cp's OEIS frontend

A306425 Smallest m such that A306305(m) = n.

Original entry on oeis.org

11, 50, 25, 14, 7, 17, 26, 13, 24, 12, 6, 3, 15, 8, 4, 2, 1, 5, 31826, 15913, 79565, 6703845

Offset: 0

Chai Wah Wu, Feb 19 2019

Sequence is finite and has only 22 terms.

Cf. A306305.

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

Original entry on oeis.org

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

V. Raman, Sep 27 2012

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

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

Cf. A045875, A215236.

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

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

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

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

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

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

A306494 Smallest number m such that n*3^m has 2 or more identical adjacent decimal digits.

Original entry on oeis.org

11, 8, 10, 8, 11, 7, 9, 5, 9, 11, 0, 7, 2, 4, 10, 2, 4, 6, 7, 8, 8, 0, 5, 4, 2, 9, 8, 4, 6, 10, 4, 2, 0, 8, 6, 6, 1, 1, 1, 8, 3, 3, 3, 0, 9, 5, 5, 1, 2, 11, 3, 7, 2, 5, 0, 7, 6, 2, 1, 7, 6, 2, 7, 5, 3, 0, 6, 4, 4, 9, 7, 3, 5, 1, 1, 1, 0, 8, 2, 5, 7, 3, 3, 3, 1

Offset: 1

Chai Wah Wu, Feb 19 2019

a(n) is smallest m such that 3^m*n is in the sequence A171901 (or -1 if no such m exists).

0 <= a(n) <= 35 for all n > 0. This is proved by showing that for each 0 < n < 10^9, there is a number m <= 35 such that 3^m*n mod 10^9 has adjacent identical digits. If n > 0 and n == 0 mod 10^9, then clearly a(n) = 0.

			a(1) = 11 since 3^11 = 177147 has 2 adjacent digits '7' and no smaller power of 3 has adjacent identical digits.
Record values:
a(1) = 11
a(241) = 12
a(2392) = 14
a(35698) = 15
a(267345) = 16
a(893521) = 17
a(29831625) = 18
a(3232453125) = 19

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A171901, A306305.

def A306494(n):
    m, k= 0, n
    while True:
        s = str(k)
        for i in range(1,len(s)):
            if s[i] == s[i-1]:
                return m
        m += 1
        k *= 3

a(A171901(n)) = 0.

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A306425 Smallest m such that A306305(m) = n.

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

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A306494 Smallest number m such that n*3^m has 2 or more identical adjacent decimal digits.

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Python

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