A302864 a(n) is the smallest positive palindromic number whose binary expansion ends in exactly n zeros, or 0 if no such number exists.

1, 2, 4, 8, 272, 25952, 2112, 4224, 8448, 44544, 2532352, 6547456, 405504, 236101632, 8634368, 487030784, 677707776, 61486268416, 25135153152, 4285005824, 2912594952192, 63115033051136, 2114151514112, 65892155129856, 29142024192, 827163020361728, 42643629092634624

Offset

0,2

Comments

Equivalently, a(n) is the smallest base-10 palindrome of the form (2k+1)*2^n.

Conjecture: a(n) > 0 for all n >= 0.

Links

Chai Wah Wu, Table of n, a(n) for n = 0..40

Formula

If A082613(n)/A082614(n) = 2^n (see conjecture at A082613) and a(n) > 0, then a(n) >= A082613(n). - Altug Alkan, Apr 15 2018

Example

   n      a(n)              binary expansion of a(n)
  == ============= ==========================================
   0             1                      1
   1             2                      10
   2             4                      100
   3             8                      1000
   4           272                  100010000
   5         25952             110010101100000
   6          2112                 100001000000
   7          4224                 1000010000000
   8          8448                 10000100000000
   9         44544                1010111000000000
  10       2532352           1001101010010000000000
  11       6547456           11000111110100000000000
  12        405504                1100011000000000000
  13     236101632        1110000100101010000000000000
  14       8634368             100000111100000000000000
  15     487030784         11101000001111000000000000000
  16     677707776         101000011001010000000000000000
  17   61486268416    111001010000110111100000000000000000
  18   25135153152      10111011010001011000000000000000000
  19    4285005824          11111111011010000000000000000000
  20 2912594952192 101010011000100100001100000000000000000000

Prog

(PARI) isp(n) = n=digits(n); for(i=1, #n\2, if(n[i]!=n[#n+1-i], return(0))); 1;
a(n) = forstep(k=1, oo, 2, if(isp(p=k*2^n), return(p))); \\ Altug Alkan, Apr 15 2018, after Charles R Greathouse IV at A002113
(Python)
from itertools import count
def A302864(n):
    for k in count(0, 2):
        if (s:=str(k+1<<n))[:(t:=(len(s)+1)//2)]==s[:-t-1:-1]:
            return k+1<<n # Chai Wah Wu, Jun 23 2022

Crossrefs

Cf. A002113, A082613, A082614.

Sequence in context: A130134 A082613 A013553 * A061089 A067499 A135212

Adjacent sequences: A302861 A302862 A302863 * A302865 A302866 A302867

Keyword

nonn,base

Author

Jon E. Schoenfield, Apr 14 2018

Extensions

a(26) from Chai Wah Wu, Mar 06 2019

Status

approved