A369127 S is a "boomerang sequence". Replace each digit d of S by a base-10 palindrome not yet used that contains d: the sequence S remains identical to itself if we follow each palindrome with a comma.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 101, 111, 202, 121, 131, 141, 151, 22, 303, 212, 161, 222, 171, 181, 33, 191, 313, 44, 414, 515, 55, 616, 232, 242, 323, 404, 333, 252, 717, 262, 818, 66, 919, 272, 282, 292, 1001, 77, 1111, 1221, 88, 1331, 343, 353, 1441, 99

Offset

1,3

Links

Michael S. Branicky, Table of n, a(n) for n = 1..10000

Example

As a(1) to a(10) are single-digit palindromes, the replacement leaves the terms a(1) to a(10) as they were.
a(11) = 11 and we must replace the first digit d = 1 of 11 by the smallest base-10 palindrome not yet used; this is 11 (as the palindrome 1 has been used before);
a(11) = 11 and we must replace now the second digit d = 1 of 11 by the smallest base-10 palindrome not yet used; this is 101 (as 1 and 11 have been used before);
a(12) = 101 and we must replace the first digit d = 1 of 101 by the smallest base-10 palindrome not yet used; this is 111;
a(12) = 101 and we must now replace the digit d = 0 of 101 by the smallest base-10 palindrome not yet used; this is 202 (as 0 and 101 have been used before);
a(12) = 101 and we must now replace the last digit d = 1 of 101 by the smallest base-10 palindrome not yet used; this is 121; etc.

Prog

(Python)
from collections import deque
from itertools import count, islice
def pals(start=1): # generator of palindromes >= palindrome start
    s = str(start)
    q, r = divmod(len(s)+1, 2)
    for d in count(q):
        olst = [1, 0][int(d==q and r==1):]
        for offset in olst:
            lb = max(1, 10**(d-1)) if d>q or offset!=olst[0] else int(s[:q])
            for i in range(lb, 10**d):
                left = str(i)
                yield int(left+left[::-1][offset:])
def agen(): # generator of terms
    S = deque([101])
    head = list(range(10)) + [11]
    yield from head
    used = set(head) | {101}
    pstart = {d:0 for d in "0123456789"}
    while True:
        an = S.popleft()
        yield an
        for d in str(an):
            p = next(p for p in pals(start=pstart[d]) if p not in used and d in str(p))
            pstart[d] = p
            S.append(p)
            used.add(p)
print(list(islice(agen(), 57))) # Michael S. Branicky, Mar 03 2024

Crossrefs

Cf. A002113, A369603, A369604, A369823, A369824, A369798.

Sequence in context: A344823 A064704 A083136 * A229623 A277856 A117057

Adjacent sequences: A369124 A369125 A369126 * A369128 A369129 A369130

Keyword

base,nonn

Author

Eric Angelini, Mar 02 2024

Extensions

More terms from Michael S. Branicky, Mar 03 2024

Status

approved