A240602 Recursive palindromes in base 2: palindromes n where each half of the digits of n is also a recursive palindrome.

0, 1, 11, 101, 111, 1111, 11011, 11111, 101101, 111111, 1010101, 1011101, 1110111, 1111111, 11111111, 111101111, 111111111, 1101111011, 1111111111, 11011011011, 11011111011, 11111011111, 11111111111, 101101101101, 111111111111, 1011010101101, 1011011101101, 1111110111111, 1111111111111

Offset

1,3

Comments

A number n with m digits in base 2 is a member of a(n) if n is a palindrome, and the first floor(m/2) digits of n is already a previous term of a(n). Fast generation of new terms with 2m digits can be done by concatenating the previous terms with m digits twice. Fast generation of new terms with 2m+1 digits can be done by concatenating the previous terms with m digits twice with any single digit in the middle. The smallest palindrome which is not a member of a(n) is 1001.

Links

Lior Manor, Table of n, a(n) for n = 1..1000

Example

11011 is in the sequence since it is a palindrome of 5 digits, and the first floor(5/2) digits of it, 11, is also a term. 1001 and 10001 are not in a(n) since 10 is not in a(n).

Mathematica

FromDigits /@ Select[IntegerDigits[Range[2^12], 2], And[PalindromeQ@ Take[#, Floor[Length[#]/2]], PalindromeQ[#]] &] (* Michael De Vlieger, Nov 08 2017 *)

Crossrefs

Cf. A057148, A240601, A165785.

Sequence in context: A076289 A350346 A247647 * A117697 A305395 A158051

Adjacent sequences: A240599 A240600 A240601 * A240603 A240604 A240605

Keyword

base,nonn

Author

Lior Manor, Apr 13 2014

Status

approved