OFFSET

1,1

COMMENTS

With the exception of 22, which is the sum of 11 and 11, no term of this sequence has an even number of digits. Proof: Other than 11, palindromes with an even number of digits are not primes (since they are divisible by 11). Suppose m is a term of this sequence with 2k digits. Then m must be the sum of two palindromic primes p and q with 2k-1 digits each. It follows that the first and the last digit of m is 1. Hence, either p or q is even, creating a contradiction with primality.

With the exception of 5, 7, and 9, all terms of this sequence are even. Proof: two consecutive multi-digit palindromes differ by at least 10, so larger palindromes can't be the sum of a palindromic prime and 2. Thus, each multi-digit term is the sum of two odd numbers.

EXAMPLE

282 can be written as the sum of two prime palindromes, 101 and 181. Thus, 282 is in the sequence.

MATHEMATICA

q := Select[Range[30000], PalindromeQ[#] && PrimeQ[#] &]

Select[Union[Flatten[Table[q[[n]] + q[[m]], {n, Length[q]}, {m, Length[q]}]]],

PalindromeQ[#] &]

PROG

(Python)

from sympy import isprime

from itertools import product

def ispal(n): s = str(n); return s == s[::-1]

def oddpals(d): # generator of odd palindromes with d digits

if d == 1: yield from [1, 3, 5, 7, 9]; return

for first in "13579":

for p in product("0123456789", repeat=(d-2)//2):

left = "".join(p); right = left[::-1]

for mid in [[""], "0123456789"][d%2]:

yield int(first + left + mid + right + first)

def auptod(dd):

N, alst, pp = 10**dd, [], [2, 3, 5, 7, 11]

pp += [p for d in range(3, dd+1, 2) for p in oddpals(d) if isprime(p)]

return sorted(set(p+q for p in pp for q in pp if p+q<N and ispal(p+q)))

print(auptod(5)) # Michael S. Branicky, Aug 29 2022

CROSSREFS

KEYWORD

nonn,base

AUTHOR

Tanya Khovanova, Aug 29 2022

STATUS

approved