A253594 Numbers n that have more than one palindromic representation in bases 2 <= b <= n-2.

10, 15, 16, 17, 18, 20, 21, 24, 26, 27, 28, 30, 31, 32, 33, 34, 36, 38, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 60, 62, 63, 64, 65, 66, 67, 68, 70, 72, 73, 74, 75, 76, 78, 80, 81, 82, 84, 85, 86, 88, 90, 91, 92, 93, 96, 98, 99, 100, 102, 104, 105, 107, 108, 109, 110, 111, 112, 114, 116, 117, 118, 119, 120, 121, 122, 123, 124

Offset: 1

Author

Christian Perfect, Jan 05 2015

Keywords

• nonn
• base
• easy

Comments

We do not include base n-1, because every n>2 is written '11' in base n-1.

Examples

			10 is written '101' in base 3, and '22' in base 4.
12 is written '22' in base 5, but is not a palindrome in any other base up to 10, so does not belong to this sequence.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1685 from Christian Perfect)

Programs

Mathematica

palbQ[b_,n_]:=With[{idb=IntegerDigits[n,b]},idb==Reverse[idb]]; Select[Range[150],Total[Boole[Table[palbQ[b,#],{b,2,#-2}]]]>1&] (* Harvey P. Dale, Aug 20 2025 *)

Python

from itertools import count
def base(n,b):
   while n:
      m = n%b
      yield m
      n = (n-m)//b
def is_palindrome(seq):
   seq = list(seq)
   l = len(seq)//2
   return seq[:l] == seq[-1:-l-1:-1]
def a():
   for n in count(2):
      base_representations = [(b,list(base(n,b))) for b in range(2,n-1)]
      pals = [(b,s) for b,s in base_representations if is_palindrome(s)]
      if len(pals)>1:
         yield n

Python

from sympy.ntheory import digits
def ok(n):
    c = 0
    for b in range(2, n-1):
        d = digits(n, b)[1:]
        c += int(d == d[::-1])
        if c == 2: return True
    return c > 1
print([k for k in range(125) if ok(k)]) # Michael S. Branicky, Feb 05 2024