OFFSET
2,1
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
FORMULA
a(A171901(n)) = 1. - Chai Wah Wu, Feb 20 2019
a(n) = A215236(n) + 1. - Georg Fischer, Nov 25 2020
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
V. Raman, Sep 27 2012
STATUS
approved