Sheldon L found out, that
$$34276387$$
is a prime factor of
$$2 \uparrow \uparrow n + 3 \uparrow \uparrow n$$
for any natural number $n \ge 5$.
$2 \uparrow \uparrow n$ leaves the remainder 20853622
$3 \uparrow \uparrow n$ leaves the remainder 13422765
- Can this result be confirmed by induction ?
- Can this result be generalized for arbitary numbers a,b ?
To make the second qeustion clearer :
Is there a natural number n for any natural numbers a,b, such that every prime factor of $a \uparrow \uparrow n+b \uparrow\uparrow n$ is a prime factor of $a \uparrow \uparrow s + b\uparrow\uparrow s$ for every $s \ge n$ ? (If so, n could be called the "stationary point")
This is a narrow answer to the "confirmation" question, showing the required steps to get a rigorous result for arbitrarily large values of n. The goal is to prove $\forall\> n>0, \> 3 \uparrow\uparrow (n+4) \bmod 34276387 = 13422765$. The key to proving the result directly for $3\uparrow\uparrow 5$ is brute force finding the optimal cycle lengths.
$\forall\> n>0, \> 3 \equiv 3^{n} \pmod 6 \\ \therefore \forall\> n>0, \> (3\uparrow\uparrow n) \bmod 6 = 3$
$\forall\> n>0, \> 3^n \equiv 3^{n+6} \pmod {156} \\ \therefore \forall\> n>0, \> 3\uparrow\uparrow (n+1) \equiv 3^{(3\uparrow\uparrow n)\bmod 6} \equiv 3^3 \equiv 27 \pmod {156}$
$\forall\> n>0, \> 3^n \equiv 3^{n+156} \pmod {94640} \\ \therefore \forall\> n>0, \> 3\uparrow\uparrow (n+2) \equiv 3^{(3\uparrow\uparrow (n+1))\bmod 156} \equiv 3^{27} \equiv 21867 \pmod {94640}$
$\forall\> n>0, \> 3^n \equiv 3^{n+94640} \pmod {34276386} \\ \therefore \forall\> n>0, \> 3\uparrow\uparrow (n+3) \equiv 3^{(3\uparrow\uparrow (n+2))\bmod 94640} \equiv 3^{21867} \equiv 7088973 \pmod {34276386}$
$3^n \equiv 3^{n+34276386} \pmod {34276387} \\ \therefore \>\forall\> n>0, \> 3\uparrow\uparrow (n+4) \equiv 3^{(3\uparrow\uparrow (n+3))\bmod 34276386} \equiv 3^{7088973} \equiv 13422765 \pmod {34276387}$
The goal is also to prove $2\uparrow\uparrow (n+4) \bmod 34276387 = 20853622 \>\forall\> n>0$. The key to proving the result directly for $2\uparrow\uparrow5$ is getting the optimal cycle lengths. At each step, of course if m>z $(z \bmod m)=z$, and in fact for $2\uparrow\uparrow 5$, the recursive steps return exactly $2\uparrow\uparrow n$. This is why the recursion result differs for $2\uparrow\uparrow 4$
$\forall\> n>0, \> 2\uparrow\uparrow (n+1) = 2^{2\uparrow\uparrow n} \>\text{and}\> 2^n \bmod 2 = 0 \\ \therefore \forall\> n>0, \> (2\uparrow\uparrow n) \bmod 2 = 0 \>\text{and}\> (2\uparrow\uparrow n) >=2 $
$\forall\> n>1, \> 2^n \equiv 2^{n+2} \pmod {12} \\ \therefore \forall\> n>0, \> 2\uparrow\uparrow (n+1) \equiv 2^{2+(2\uparrow\uparrow n)\bmod 2} \equiv 2^2 \equiv 4 \pmod {12}$
$\forall\> n>2, \> 2^n \equiv 2^{n+12} \pmod {3640} \\ \therefore \forall\> n>0, \> 2\uparrow\uparrow (n+2) \equiv 2^{(2\uparrow\uparrow (n+1))\bmod 12} \equiv 2^4 \equiv 16 \pmod {3640}$
$\forall\> n>0, \> 2^n \equiv 2^{n+3640} \pmod {34276386} \\ \therefore \forall\> n>0, \> 2\uparrow\uparrow (n+3) \equiv 2^{(3\uparrow\uparrow (n+2))\bmod 3640} \equiv 2^{16} \equiv 65536 \pmod {34276386}$
$2^n \equiv 2^{n+34276386} \pmod {34276387} \\ \therefore \forall\> n>0, \> 2\uparrow\uparrow (n+4) \equiv 2^{(2\uparrow\uparrow (n+3))\bmod 34276386} \equiv 2^{65536} \equiv 20853622 \pmod {34276387}$
Finally, $13422765+20853622=34276387 \therefore \> \forall\>n>4,$ $$(3\uparrow\uparrow n) + (2\uparrow\uparrow n) \equiv 0 \pmod {34276387}$$
By contrast, the pari-gp program, from my answer to a previous question, using the unoptimized chain length, always iterating $m\mapsto\phi(m)$ gets the same result, and can give results to arbitrary tetration heights, but the "proof" would involve checking all tetration heights<=24.