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A356854
Palindromes that can be written in more than one way as the sum of two distinct palindromic primes.
1
282, 484, 858, 888, 21912, 22722, 23832, 24642, 25752, 26662, 26762, 26862, 26962, 27672, 27772, 27872, 27972, 28482, 28782, 28882, 28982, 29692, 29792, 29892, 29992, 40704, 41514, 41614, 41814, 42624, 42824, 42924, 43434, 43734, 43834, 43934, 44744, 44844, 44944, 45354
OFFSET
1,1
COMMENTS
This sequence doesn't contain any numbers with an even number of digits, see proof in A356824.
Subsequence of A356824.
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.
EXAMPLE
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.
MATHEMATICA
q := Select[Range[50000], PalindromeQ[#] && PrimeQ[#] &]
Sort[Transpose[Select[Tally[Flatten[Table[q[[n]] + q[[m]], {n, Length[q]}, {m, n + 1, Length[q]}]]], PalindromeQ[#[[1]]] && #[[2]] > 1 &]][[1]]]
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, once, twice = 10**dd, [], [2, 3, 5, 7, 11], set(), set()
pp += [p for d in range(3, dd+1, 2) for p in oddpals(d) if isprime(p)]
sums = (p+q for p in pp for q in pp if p<q and p+q<N and ispal(p+q))
for s in sums:
if s in once: twice.add(s)
else: once.add(s)
return sorted(twice)
print(auptod(5)) # Michael S. Branicky, Aug 31 2022
CROSSREFS
Cf. A356824.
Sequence in context: A242982 A255387 A320706 * A114804 A250580 A252364
KEYWORD
nonn,base
AUTHOR
STATUS
approved