OFFSET
1,1
COMMENTS
It is conjectured that are just 3 palindromic primes with digit sum 2, namely 2, 11 and 101. If any others exist, they must be of the form 10^(2^k) + 1 with k > 14.
From Jeppe Stig Nielsen, Aug 30 2025: (Start)
It is now known that any additional primes 10^(2^k) + 1 must have k >= 31.
Digit sum 3 yields only one prime, 3, a palindrome in a vacuous way.
Digit sum 4 leads to primes (A062339), but such numbers can never be palindromes. Proof: Let w be any palindrome with digit sum 4. So w = 10^a + 10^b + 10^c + 10^d with a >= b >= c >= d >= 0. But then 10^c + 10^d is a nontrivial divisor of w, showing that w is not prime.
You may have come here searching for the subsequence 5, 131, 10301, 1003001, 100030001, 10000000000300000000001, ... where the largest digit exceeds 1. See A171376 and A100028 for information on them.
(End)
Several terms are known that have three 1s in the middle. See A344424. - Jeppe Stig Nielsen, Dec 14 2025
LINKS
Jeppe Stig Nielsen, Table of n, a(n) for n = 1..386 (all terms below 10^1000; terms n = 1..238 from Chai Wah Wu)
Hans Riesel, Some factors of the numbers Gn = 6^2^n+1 and Hn = 10^2^n+1, Math. Comp. 23 (1969), p. 413-415. With errata reported in Math. Comp. 24 (1970), p. 243.
MATHEMATICA
Do[p = Join[ IntegerDigits[n, 4], Reverse[ Drop[ IntegerDigits[n, 4], -1]]]; q = Plus @@ p; If[q == 5 && PrimeQ[ FromDigits[p]] && q == 5, Print[ FromDigits[p]]], {n, 1, 4 10^8}] (* this coding will not pick up the first entry *)
PROG
(PARI) for(i=0, 50, for(j=0, i, p=10^(2*i)+10^(i+j)+10^i+10^(i-j)+1; isprime(p)&&print1(p, ", "))) \\ Jeppe Stig Nielsen, Aug 30 2025
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Amarnath Murthy, May 05 2002
EXTENSIONS
Edited by Robert G. Wilson v, May 15 2002
More terms from Chai Wah Wu, Nov 25 2015
STATUS
approved
