A344550 Nested palindromes.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 111, 222, 333, 444, 555, 666, 777, 888, 999, 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999, 10101, 11111, 12121, 13131, 14141, 15151, 16161, 17171, 18181, 19191, 20202, 21212, 22222, 23232, 24242

Offset: 1

Author

James S. DeArmon, May 22 2021

Keywords

• nonn
• base

Comments

Both the right and left halves of each term are themselves palindromes.

Here, "half" means ceiling(m/2) digits for an m-digit term, whereas A240601 uses floor(m/2). - Michael S. Branicky, May 22 2021

Examples

			2222 is a nested palindrome since 22=22, and taking just one side, 2=2. For an example using an odd number of digits, 68686 is a nested palindrome since 686 is a palindrome. Using parentheses to indicate nesting: ( (22) (22) ) and ( (686) (686) ), where, in the second example, the middle-most 6 is repeated for exposition.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..250

Crossrefs

Cf. A002113, A240601.

Programs

Mathematica

Select[Range[0,10^5],(k=#;f=FromDigits/@(Take[IntegerDigits[k],#]&/@{l=Ceiling[IntegerLength@k/2],-l});
And@@PalindromeQ/@Join[f,{k}])&] (* Giorgos Kalogeropoulos, Jun 24 2021 *)
Select[Range[0,25000],AllTrue[{#,FromDigits[Take[IntegerDigits[#],Ceiling[ IntegerLength[ #]/2]]]},PalindromeQ]&] (* Harvey P. Dale, May 15 2022 *)

Perl

foreach $cand (0..200000){
    @a=split("",$cand);
    $b = join("",reverse @a);
    next unless $cand==$b; # palindromes only
    $len = int(@a/2.); $lenA = @a;
    $len++ unless ($lenA/2 == int $lenA/2); # an even half? or include middle digit
    $half = join("",@a[0..($len-1)]);
    $revHalf = join("",reverse @a[0..($len-1)]);
    next unless $half == $revHalf;
    $str .= "$cand, ";
}
chop $str;  chop $str;  # remove trailing comma and space
print "$str\n";  # write to stdout

Python

from itertools import product
def pals(d, base=10): # returns a string
  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]:
        if left + mid + right != '0': yield left + mid + right
def auptod(dd):
  yield 0
  for d in range(1, dd+1):
    yield from (int(p+p[-1-d%2::-1]) for p in pals((d+1)//2))
print([np for np in auptod(6)]) # Michael S. Branicky, May 22 2021