What can we say about the prime factors of $​^{10}10+23$? In a video on ultrafinitism I saw a claim that the number $​^{10}10+23$ does not have prime factorization. While I don't accept the premise of ultrafinitism, I got curious, what can we say about the prime factors of this number?
$​^{10}10$ refers to the hyperoperation tetration. In other words, the number is equal to $10^{10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}}$, that is an exponent tower of $10$ that is $10$ high or $10\uparrow\uparrow10$ in the Knuth's up-arrow notation.
Obviously, we can say that $​^{10}10+23$ is not divisible by $2$ or $5$. Furthermore, we can say that this number is divisible by $3$ but only once, since the digits of this number add up to $6$.
What else can we say about the prime factors of this number?
 A: I 'll try to give a limited answer to the title question.
First : $N_1 = 111111$ ( $6$ digits ) is divisible by 13. Then any number m such $m = k \times N_1$ is also divisible by 13. 
Let's consider a number $f(p) = 10^p + 23$ with $p = 6q+4$. For $q=0$ we get $100023$ which is divisible by $13$.
Let's assume that it is true for q and show it is true for q'=q+1. A way to show it is to check the difference between them. If we add a multiple of $13$ to a multiple of $13$, the sum is also a multiple of $13$ :
$f(p+6)-f(p) = 10^{6q+6+4} + 23 - 10^{6q+4} - 23 = 10^{6q+4} \times (10^{6}-1) = 999999 \times 10^{6q+4} = 3^2 \times 111111 \times  10^{6q+4}$. We know from the preliminary that $111111$ is a multiple of $13$.
If it is true for q=0 and q true => q+1 true, we may conclude that any power of 10 of the form $6k+4$ leads to a number divisible by 13.
We know also that the power of the $10$ of the question expression  $ 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} \pmod 6 \equiv 4$. Bingo, it is the condition for the 13 divisibility. Then our meta number is not the triple of a prime
Similar methods may be used to show that $​^{10}10+23$ is not divisible by 11 which needs the power to be odd and nor by $41$ which needs a not fulfilled power of the form $5 k + 2$. Credit to Fleablood which felt the 13 ... I searched first for the recurrent $11$.
More complex regularities complete these ones. Sometime $\frac{10^n + 23}{3}$ is a prime ( f.e. $n=6$) and it is remarkable that in general for any $n$ there is a very few factors.
This doesn't disallow the controversial proposal of non-existence of such numbers, assuming that is a consistent proposal. For my own, these considerations have nothing to do in maths, even if similar postures are useful in physics. Maths aren't a field of existence, imagination has not limit ...
Edit : I found also that our number is :


*

*not divisible by $7$ : $6k+5$ instead of $6k+4$

*not divisible by $19$ : $18k+7$ instead of $18k+10$

*not divisible by $31$ : $15k+3$ instead of $15k+10$

A: As shown here, the known prime factors are 3, 13, 673, 18301, 400109, and 27997373.
