A240601 Recursive palindromes in base 10: palindromes n where each half of the digits of n is also a recursive palindrome.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 505, 515, 525, 535, 545, 555, 565, 575, 585, 595, 606, 616, 626, 636, 646, 656, 666, 676, 686, 696, 707, 717, 727, 737, 747, 757, 767, 777, 787, 797, 808, 818, 828, 838, 848, 858, 868, 878, 888, 898, 909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1111

Offset: 1

Author

Lior Manor, Apr 09 2014

Keywords

• base
• nonn
• nice

Comments

A number n with m digits in base 10 is a member if n is a palindrome, and the first floor(m/2) digits of n is already a previous term of a(n). All repdigit numbers are terms 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.

Examples

			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 the sequence since 10 is not in the sequence.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Lior Manor)

Crossrefs

Cf. A002113, A010785, A240602.

Cf. A227858 (first difference is a(110) = 1111, but A227858(109) = 1001). - Georg Fischer, Oct 23 2018

Programs

Python

from itertools import product
def pals(d, base=10): # all d-digit palindromes as strings
  digits = "".join(str(i) for i in range(base))
  for p in product(digits, repeat=d//2):
    if d//2 > 0 and p[0] == "0": continue
    left = "".join(p); right = left[::-1]
    for mid in [[""], digits][d%2]: yield left + mid + right
def auptod(dd):
  for d in range(1, dd+1):
    for p in pals(d//2):
      if d//2 == 0: p = ""
      elif p[0] == "0": continue
      for mid in [[""], "0123456789"][d%2]: yield int(p+mid+p[::-1])
print([rp for rp in auptod(6)]) # Michael S. Branicky, May 22 2021