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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.)