A355404 Lexicographically earliest sequence of distinct terms such that the concatenation of three successive terms form a palindrome using the alphabet {1, 2}.

1, 2, 21, 22, 12, 122, 121, 221, 22121, 22122, 12122, 12122122, 12122121, 22122121, 2212212122121, 2212212122122, 1212212122122, 121221212212212122122, 121221212212212122121, 221221212212212122121, 2212212122122121221212212212122121

Offset

1,2

Comments

Also, lexicographically earliest sequence of distinct terms > 0 such that the concatenation of three successive terms form a palindrome in bases >= 3.

a(45) with 1597 digits is the first term > 10^1000.

Links

Michael S. Branicky, Table of n, a(n) for n = 1..44

Prog

(Python)
from itertools import count, islice, product
def pals(digs):
    yield from digs
    for d in count(2):
        for p in product(digs, repeat=d//2):
            left = "".join(p)
            for mid in [[""], digs][d%2]:
                yield left + mid + left[::-1]
def folds(s): # generator of suffixes of palindromes starting with s
    for i in range((len(s)+1)//2, len(s)+1):
        for mid in [True, False]:
            t = s[:i] + (s[:i-1][::-1] if mid else s[:i][::-1])
            if t.startswith(s):
                yield t[len(s):]
    yield from ("".join(p)+s[::-1] for p in pals("12"))
def agen():
    s, t, seen = "1", "2", {"1", "2"}
    yield from [1, 2]
    while True:
        for u in folds(s+t):
            if len(u) > 0 and u not in seen: break
        s, t = t, u
        seen.add(u)
        yield int(t)
print(list(islice(agen(), 21))) # updated Michael S. Branicky, Aug 09 2022

Crossrefs

Cf. A002113, A091789.

Sequence in context: A320523 A371060 A171549 * A077678 A037309 A037315

Adjacent sequences: A355401 A355402 A355403 * A355405 A355406 A355407

Keyword

nonn,base

Author

Michael S. Branicky, Jul 01 2022

Status

approved