The minimum sum over all digits of palindrome primes is 2,
excluding single digit primes, which occurs for the primes 11 and
101. It is conjectured that there are no other palprimes with a sum
over all digits of 2 and it is proved that all palindrome numbers
with a sum over all digits of 3 and 4 are not prime. The next
smallest sum over all digits is 5 for which there are two cases
which exhibit primes and are explored below.

The palindromes N₂[m] = 10^m + 1 have no primes for m > 2.

Conjecture and partial proof: Since

it can be concluded that N₂[2m + 1] = 10^2m+1 + 1 contains no primes for m > 0.
For even values of m, N₂[2m], can be written as N₂[k,n] = 10^2k(2n+1) + 1
where by the argument above N₂[k,n] contains no primes for n > 0
which leaves the n = 0 case, D[k] = 10^2k + 1 as the only possible
numbers which might be prime. Note that D(k) is the greatest
common denominator (GCD) of N₂[k,n] for n > 0.

The lack of primality of D(k) was tested by calculating the smallest
denominator (SD) of D(k) by use of "Mathematica's" IntegerQ[n] function
on the ratio D(k)/Prime[m] with m increasing from 1 to until IntegerQ[ratio]
was True. The result was checked by dividing SD into D(k) and using IntegerQ[n]
on the result. The results for the SD's are displayed in Table 1. This means
that the smallest possible prime for D(k) with k > 1 must have at least 2097152
digits (k = 21).

Palindromes with a sum over all digits equal to 3 are not prime.

Proof: The palindrome numbers N₃[m] = 10^2m + 10^m + 1 have a GCD of 3
for m = 1 through 3. Assuming that N₃[m] is divisible by 3 then it
can be shown that N₃[m + 1]/3 = {100*N₃[m] – 9*10^m+1 – 99}/3 so that there
are no palindrome primes with a sum over all digits of 3.

Palindromes with a sum over all digits equal to 4 are not prime.

Proof: Palindrome numbers with a sum over all digits of 4 can be
written as N₄[n,m] = 10^m+n + 10^m + 10ⁿ + 1 = (10^m +1)(10ⁿ + 1) which if the
number of digits in the number, n + m + 1, are fixed then m and n
are varied over all values with m > n such that their sum is m + n.
The other possible number, N₄[m] = 10^2m + 2*10^m + 1, is a perfect square.
Hence there are no palindrome primes with a sum over all digits of 4.

For the sum over all digits of 5 there are two possible types of palindrome numbers.

PRIME SEARCHES FOR PRIMES OF THE GENERALIZED FERMAT NUMBERS OF BASE 10

N(a, n) = (a^(2^n)) + 1

with a an integer it can be shown that

N(a, n+k) = (N(a, n) – 1)^(2^k) + 1

Note that if a is an odd integer N(a, n) must be even and hence all N(a, n)
are divisible by 2 and no primes are allowed. When a is an even integer N(a, n)
is odd and each N(a, n) has the curious property that each N(a,n) has a unique
set of divisors. In other words, if D is a divisor of N(a,n) then it cannot be
a divisor of N(a,n+k) for all integers k > 0. This does not foreclose on these
numbers being prime. In the case of generalized Fermat numbers of base 10 the
prime search situation is summarized below in Table 1.