OFFSET
1,1
COMMENTS
Equivalently, semiprime palindromes where both prime factors are palindromes. - Franklin T. Adams-Watters, Apr 11 2011
The sequence "trivially" includes products of palindromic primes p*q where
a) p = 2 or 3 and q has only digits < 4, as q = 11, 101, 131, 10301, 30103, ...
b) p <= 11 and q has only digits 0 and 1, as q = 101 and repunit primes A004022
c) p = 11 and q has only digits spaced out by zeros, as q = 101, 10301, 10501, 10601, 30103, 30203, 30403, 30703, 30803, ... - M. F. Hasler, Jan 04 2022
LINKS
Lars Blomberg, Table of n, a(n) for n = 1..1000
FORMULA
EXAMPLE
The palindrome 35653 is a term since it has 2 factors, 101 and 353, both palindromic.
MATHEMATICA
Take[Select[Times@@@Tuples[Select[Prime[Range[5000]], PalindromeQ], 2], PalindromeQ]// Union, 50] (* Harvey P. Dale, Aug 25 2019 *)
PROG
(PARI) {first(N=50, p=1) = vector(N, i, until( bigomega( p=nxt_A002113(p))==2 && vecmin( apply( is_A002113, factor(p)[, 1])), ); p)} \\ M. F. Hasler, Jan 04 2022
(Python)
from sympy import factorint
from itertools import product
def ispal(n): s = str(n); return s == s[::-1]
def pals(d, base=10): # all d-digit palindromes
digits = "".join(str(i) for i in range(base))
for p in product(digits, repeat=d//2):
if d > 1 and p[0] == "0": continue
left = "".join(p); right = left[::-1]
for mid in [[""], digits][d%2]: yield int(left + mid + right)
def ok(pal):
f = factorint(pal)
return sum(f.values()) == 2 and all(ispal(p) for p in f)
print(list(filter(ok, (p for d in range(1, 6) for p in pals(d) if ok(p))))) # Michael S. Branicky, Aug 14 2022
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Patrick De Geest, Jun 15 1998
EXTENSIONS
Definition clarified by Franklin T. Adams-Watters, Apr 11 2011
More terms from Lars Blomberg, Nov 06 2015
STATUS
approved