A113612 - OEIS

%I #21 Aug 13 2022 05:16:05

%S 3,13,313,13313,631331313313,10131331313313631331313313,

%T 1331363133131331310131331313313631331313313

%N 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

%C From _Michael S. Branicky_, Aug 12 2022: (Start)

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

%C a(12) has 1528 digits. (End)

%H Michael S. Branicky, <a href="/A113612/b113612.txt">Table of n, a(n) for n = 1..11</a>

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

%o (Python)

%o from sympy import isprime

%o from itertools import count, islice, product

%o def pals(digs):

%o     yield from digs

%o     for d in count(2):

%o         for p in product(digs, repeat=d//2):

%o             left = "".join(p)

%o             for mid in [[""], digs][d%2]:

%o                 yield left + mid + left[::-1]

%o def folds(s): # generator of suffixes of palindromes starting with s

%o     for i in range((len(s)+1)//2, len(s)+1):

%o         for mid in [True, False]:

%o             t = s[:i] + (s[:i-1][::-1] if mid else s[:i][::-1])

%o             if t.startswith(s):

%o                 yield t[len(s):]

%o     yield from ("".join(p)+s[::-1] for p in pals("0123456789"))

%o def agen():

%o     s, seen = "3", {"3"}; yield from [3]

%o     while True:

%o         for t in folds(s):

%o             if len(t) and t[0] != "0" and t not in seen and isprime(int(t)):

%o                 break

%o         s += t; seen.add(t); yield int(t)

%o print(list(islice(agen(), 7))) # _Michael S. Branicky_, Aug 12 2022

%Y Cf. A113613.

%K nonn,base

%O 1,1

%A _Amarnath Murthy_, Nov 09 2005

%E Definition clarified by _Felix Fröhlich_, Oct 27 2014

%E Name clarified, a(2)-a(4) corrected, and a(5) and beyond from _Michael S. Branicky_, Aug 12 2022