A369233 Smallest base for which the digits expansion of 2^n is palindromic.

3, 3, 3, 3, 7, 7, 7, 15, 7, 7, 31, 7, 15, 15, 31, 15, 15, 31, 63, 15, 31, 31, 127, 63, 31, 31, 63, 127, 127, 31, 63, 63, 255, 255, 127, 63, 63, 127, 511, 255, 255, 63, 127, 127, 511, 511, 511, 255, 127, 127, 255, 1023, 1023, 511, 511, 127, 255, 255, 2047, 1023, 1023, 1023, 127, 255

Offset: 1

Author

Michel Marcus, Jan 17 2024

Keywords

• nonn
• base
• look

Comments

From the Kreher and Stinson article we know that a(n) is of the form 2^k-1 (cf. A000225). - David A. Corneth, Jan 18 2024

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000 (first 400 terms from Michel Marcus)
• Donald L. Kreher and Douglas R. Stinson, On min-base palindromic representations of powers of 2, arXiv:2401.07351 [math.NT], 2024. See Table 3 p. 7.

Crossrefs

Cf. A000079, A000225, A016026.

Programs

Maple

f:= proc(n) local x,b,L,i;
  x:= 2^n;
  for b from 3 do
    L:= convert(x,base,b);
    if andmap(i -> L[i]=L[-i], [$1..nops(L)/2]) then return b fi
  od
end proc:
map(f,[$1..100]); # Robert Israel, Jan 17 2024

Mathematica

A369233[n_] := Block[{p = 2^n, k = 1}, While[!PalindromeQ[IntegerDigits[p, 2^++k-1]]]; 2^k-1]; Array[A369233, 100] (* Paolo Xausa, Mar 10 2024 *)

PARI

ispal(n, b) = my(d=digits(n, b)); d == Vecrev(d);
a(n) = my(b=2, N=2^n); while (! ispal(N, b), b++); b;

PARI

a(n) = {my(pow2 = 1<David A. Corneth, Jan 18 2024

Python

from itertools import count
from sympy.ntheory.factor_ import digits
def A369233(n):
    m = 1<>1)]==s[:-t-1:-1]) # Chai Wah Wu, Jan 17 2024

Formula

a(n) = A016026(A000079(n)).