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

%I A240601 #20 May 22 2021 20:39:49
%S A240601 0,1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99,101,111,121,131,141,
%T A240601 151,161,171,181,191,202,212,222,232,242,252,262,272,282,292,303,313,
%U A240601 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
%N A240601 Recursive palindromes in base 10: palindromes n where each half of the digits of n is also a recursive palindrome.
%C A240601 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.
%H A240601 Michael S. Branicky, <a href="/A240601/b240601.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from Lior Manor)
%e A240601 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.
%o A240601 (Python)
%o A240601 from itertools import product
%o A240601 def pals(d, base=10): # all d-digit palindromes as strings
%o A240601   digits = "".join(str(i) for i in range(base))
%o A240601   for p in product(digits, repeat=d//2):
%o A240601     if d//2 > 0 and p[0] == "0": continue
%o A240601     left = "".join(p); right = left[::-1]
%o A240601     for mid in [[""], digits][d%2]: yield left + mid + right
%o A240601 def auptod(dd):
%o A240601   for d in range(1, dd+1):
%o A240601     for p in pals(d//2):
%o A240601       if d//2 == 0: p = ""
%o A240601       elif p[0] == "0": continue
%o A240601       for mid in [[""], "0123456789"][d%2]: yield int(p+mid+p[::-1])
%o A240601 print([rp for rp in auptod(6)]) # _Michael S. Branicky_, May 22 2021
%Y A240601 Cf. A002113, A010785, A240602.
%Y A240601 Cf. A227858 (first difference is a(110) = 1111, but A227858(109) = 1001). - _Georg Fischer_, Oct 23 2018
%K A240601 base,nonn,nice
%O A240601 1,3
%A A240601 _Lior Manor_, Apr 09 2014