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A258757 For 1 < k <= n, let m be the largest number such that k^1, k^2, ... k^m are palindromes in base n. a(n) gives the smallest k which has the largest value of m. 1
2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 6, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 7, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 10, 2, 2, 2, 2, 2, 2, 2, 6, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 11, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET

2,1

COMMENTS

Excluding repetitions, this is a permutation of the natural numbers excluding all perfect powers (i.e., this sequence contains N if and only if N is in A007916). - Derek Orr, Jun 18 2015

LINKS

Christian Perfect, Table of n, a(n) for n = 2..1000

EXAMPLE

a(8) = 3 because the first 6 powers of 3 are palindromes in base 8, which is more than any other number in the range 2..8 (here, m = 6).

PROG

(Python)

def to_base(n, b):

...s = []

...while n:

......m = n%b

......s = [m]+s

......n = (n-m)//b

...return s

.

def is_palindrome(n):

...return n==n[::-1]

.

def num_palindromes(n, b):

...if n<2:

......return 0

...t = 1

...while is_palindrome(to_base(n**t, b)):

......t+=1

...return t-1

.

def most_palindromes(b):

...return max(range(2, b+1), key=lambda n:num_palindromes(n, b))

(PARI) a(n)=v=[-1]; for(k=2, n, i=1; c=0; while(i, d=digits(k^i, n); if(Vecrev(d)==d, c++); if(Vecrev(d)!=d, break); i++); if(c>v[#v], v=concat(v, c); m=k)); m

vector(100, n, a(n+1)) \\ Derek Orr, Jun 18 2015

CROSSREFS

Cf. A007916.

Sequence in context: A184156 A187785 A238277 * A024708 A096917 A335108

Adjacent sequences:  A258754 A258755 A258756 * A258758 A258759 A258760

KEYWORD

nonn,base

AUTHOR

Christian Perfect, Jun 09 2015

STATUS

approved

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Last modified August 10 12:39 EDT 2020. Contains 336379 sequences. (Running on oeis4.)