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.

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

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 * A351399 A024708 A096917

Adjacent sequences: A258754 A258755 A258756 * A258758 A258759 A258760

Keyword

nonn,base

Author

Christian Perfect, Jun 09 2015

Status

approved