%I #11 Sep 04 2022 12:46:38
%S 282,484,858,888,21912,22722,23832,24642,25752,26662,26762,26862,
%T 26962,27672,27772,27872,27972,28482,28782,28882,28982,29692,29792,
%U 29892,29992,40704,41514,41614,41814,42624,42824,42924,43434,43734,43834,43934,44744,44844,44944,45354
%N Palindromes that can be written in more than one way as the sum of two distinct palindromic primes.
%C This sequence doesn't contain any numbers with an even number of digits, see proof in A356824.
%C Subsequence of A356824.
%C All numbers in this sequence are even. Proof: any 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 term is the sum of two odd numbers.
%e 282 can be expressed as a sum of two distinct palindromic primes in two ways: 282 = 101 + 181 = 131 + 151. Thus, 282 is in this sequence.
%t q := Select[Range[50000], PalindromeQ[#] && PrimeQ[#] &]
%t Sort[Transpose[Select[Tally[Flatten[Table[q[[n]] + q[[m]], {n, Length[q]}, {m, n + 1, Length[q]}]]], PalindromeQ[#[[1]]] && #[[2]] > 1 &]][[1]]]
%o (Python)
%o from sympy import isprime
%o from itertools import product
%o def ispal(n): s = str(n); return s == s[::-1]
%o def oddpals(d): # generator of odd palindromes with d digits
%o if d == 1: yield from [1, 3, 5, 7, 9]; return
%o for first in "13579":
%o for p in product("0123456789", repeat=(d-2)//2):
%o left = "".join(p); right = left[::-1]
%o for mid in [[""], "0123456789"][d%2]:
%o yield int(first + left + mid + right + first)
%o def auptod(dd):
%o N, alst, pp, once, twice = 10**dd, [], [2, 3, 5, 7, 11], set(), set()
%o pp += [p for d in range(3, dd+1, 2) for p in oddpals(d) if isprime(p)]
%o sums = (p+q for p in pp for q in pp if p<q and p+q<N and ispal(p+q))
%o for s in sums:
%o if s in once: twice.add(s)
%o else: once.add(s)
%o return sorted(twice)
%o print(auptod(5)) # _Michael S. Branicky_, Aug 31 2022
%Y Cf. A356824.
%K nonn,base
%O 1,1
%A _Tanya Khovanova_ and _Massimo Kofler_, Aug 31 2022