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 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 2
 3, 13, 313, 13313, 631331313313, 10131331313313631331313313, 1331363133131331310131331313313631331313313 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified September 22 01:06 EDT 2023. Contains 365503 sequences. (Running on oeis4.)