A113612 a(1) = 3. For n > 1, a(n) = smallest prime not already seen forming a palindrome when a(n) and all previous terms are concatenated

3, 13, 313, 13313, 631331313313, 10131331313313631331313313, 1331363133131331310131331313313631331313313

Offset

1,1

Comments

From Michael S. Branicky, Aug 12 2022: (Start)

If terms are not constrained to be distinct, then 3, 3, 3, 3, ... and 3, 13, 13, 13, ... would be solutions.

a(12) has 1528 digits. (End)

Links

Michael S. Branicky, Table of n, a(n) for n = 1..11

Example

3, 313, 313313, 31331313313, ... are all palindromes.

Prog

(Python)
from sympy import isprime
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("0123456789"))
def agen():
    s, seen = "3", {"3"}; yield from [3]
    while True:
        for t in folds(s):
            if len(t) and t[0] != "0" and t not in seen and isprime(int(t)):
                break
        s += t; seen.add(t); yield int(t)
print(list(islice(agen(), 7))) # Michael S. Branicky, Aug 12 2022

Crossrefs

Cf. A113613.

Sequence in context: A000859 A045748 A113526 * A180654 A260576 A156358

Adjacent sequences: A113609 A113610 A113611 * A113613 A113614 A113615

Keyword

nonn,base

Author

Amarnath Murthy, Nov 09 2005

Extensions

Definition clarified by Felix Fröhlich, Oct 27 2014

Name clarified, a(2)-a(4) corrected, and a(5) and beyond from Michael S. Branicky, Aug 12 2022

Status

approved