A113613 Beginning with 7, distinct primes such that every partial concatenation is a palindrome.

7, 17, 5717, 27175717, 212717571727175717, 8717571727175717212717571727175717, 2327175717271757172127175717271757178717571727175717212717571727175717

Offset

1,1

Comments

From Michael S. Branicky, Aug 09 2022: (Start)

If terms were not required to be distinct, then 7, 7, 7, ... or 7, 17, 17, 17, ... satisfy the requirement.

a(11) has 1132 digits. (End)

Links

Michael S. Branicky, Table of n, a(n) for n = 1..10

Example

7, 717, 7175717, 717571727175717, ... 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 = "7", {"7"}; yield 7
    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 09 2022

Crossrefs

Cf. A113612.

Sequence in context: A177366 A138491 A022511 * A070415 A034083 A185455

Adjacent sequences: A113610 A113611 A113612 * A113614 A113615 A113616

Keyword

base,nonn

Author

Amarnath Murthy, Nov 09 2005

Extensions

Name clarified and a(5) and beyond from Michael S. Branicky, Aug 09 2022

Status

approved