This is not a complete answer, just my observations.
In short, I have split the "simpler" recursion into "decreasing" and "short" patterns.
For the initial recursion, I have only computations as everything seems much more chaotic.
The definitions and question
Let $(a,b)$ and $(a,b,c)$ be starting conditions for your recursions (for $n=1,2$ and $n=1,2,3$).
We write $f_n=f_n(a,b)$ for the recursion $f_n=f_{n-1}f_{n-2} \bmod[f_{n-1}+f_{n-2}],n\gt 2$.
We write $f_n=f_n(a,b,c)$ for $f_n=f_{n-1}f_{n-2}f_{n-3} \bmod[f_{n-1}+f_{n-2}+f_{n-3}],n\gt 3$.
If there exists $n_0$ such that $\forall n\ge n_0,f_n\in F$, we write $f_n\to F$, where $F$ is a tuple (ordered set) that represents a cycle. In this case, we say that $f_n$ converges to $F$.
Additionally, specially define $0\pmod 0:=0$ so "consecutive zeroes" (sequence terminates) can now be treated as a cycle $F_0=(0)=0$ (if $F$ has only one element, we can write it as a number instead).
Your question now becomes:
Does $f_n$ always converge to some cycle $F$?
If for some $n_0$ we have that $\forall n\ge n_0$, $f_n(a_1,b_1)= f_{n-n_0+1}(a_2,b_2)$, then we say that the pattern (starting conditions) $(a_1,b_1)$ converge to the pattern (sequence given by) $(a_2,b_2)$.
So we have to prove that all sequences either converges to some $F$ or to some other converging sequence.
About the "simpler" $f_n(a,b)$ recursive function
Proving that the "simpler" recursive sequence $f_n=f_n(a,b)$ always terminates looks possible, but hard.
I claim that, every pair of starting conditions $(a,b)$ either converges into the "decreasing pattern", or converges in finitely many steps following one of the "short patterns".
The "decreasing pattern" are sequences that could be extended to have arbitrarily large $n_0$, but still converge to some $F$. Otherwise, we have the "short pattern" of sequences that converge in at most $n_0\le n_0^{\text{max}}$ steps, for some constant $n_0^{\text{max}}$.
I claim the "decreasing pattern" is given by these three families of starting conditions:
$$\begin{array}{}
f_n(6k+0,6k-6)\to 0, & n_0=k+1, & \text{(Ends as: $...,12,6,0.$)}\\
f_n(6k+2,6k-4)\to 0, & n_0=k+5, & \text{(Ends as: $...,14,8,2,6,4,4,0.$)}\\
f_n(6k+4,6k-2)\to 0, & n_0=k+3, & \text{(Ends as: $...,16,10,4,12,0.$)}\\
\end{array}$$
Where $k\ge2$ is a positive integer.
In other words, I claim that $n_0$ can be arbitrarily large if and only if the sequence $f_n(a,b)$ belongs to the "decreasing pattern" or converges into it. Otherwise, it either belongs to or converges to one of the "short patterns" in at most $n_0^{\text{max}}\lt\infty$ steps.
This claim would imply that $f_n(a,b)$ always converges in finitely many steps $n_0$.
This claim was verified for all possible pairs $(a,b)$ such that $a,b\le 2000$, so far.
The record breakers for the longest "short patterns", that have been observed so far, are:
$$\begin{array}{}
f_n(1,1) & \to1, & n_0=2\\
f_n(1,2) & \to0, & n_0=4\\
f_n(2,1) & \to0, & n_0=5\\
f_n(3,2) & \to0, & n_0=6\\
f_n(3,4) & \to3, & n_0=7\\
f_n(2,9) & \to95, & n_0=14\\
f_n(11,2) & \to95, & n_0=15\\
f_n(12,19) & \to\{7,11,5\}, & n_0=17\\
f_n(21,8) & \to\{7,11,5\}, & n_0=20\\
f_n(24,23) & \to\{7,11,5\}, & n_0=21\\
f_n(16,27) & \to15, & n_0=23\\
f_n(29,13) & \to\{7,11,5\}, & n_0=25\\
f_n(7,32) & \to\{7,11,5\}, & n_0=27\\
f_n(28,37) & \to\{7,11,5\}, & n_0=38\\
f_n(9,52) & \to855, & n_0=59\\
f_n(57,124) & \to855, & n_0=61\\
f_n(126,113) & \to855, & n_0=77\\
f_n(145,126) & \to855, & n_0=78\\
f_n(305,261) & \to855, & n_0=79\\
f_n(948,889) & \to455, & n_0=80\\
f_n(350,1073) & \to855, & n_0=81\\
f_n(1159,1106) & \to6399, & n_0=85\\
f_n(157,1241) & \to8775, & n_0=93\\
f_n(942,1387) & \to54675, & n_0=99\\
&\dots&
\end{array}$$
That is, so far, $n_0^{\text{max}}\ge 99$.
A potential issue could be that, the "decreasing pattern" is incomplete.
That is, are there more sequences that could have arbitrarily large $n_0(k)\gt n_0^{\text{max}}$, other than sequences that converge into one of the three families defined under "decreasing pattern"?
Assuming no such issue, the main problem is characterizing all of the "short patterns", which looks hard.
First, here are some easy conclusions to start with:
- We can assume $a\ne b$ since it is not hard to see that:
$$
f_n(a,a)\to\begin{cases}0 & (n_0=3), & 2\mid a \\a & (n_0=1), & 2\not\mid a \end{cases}
$$
- We can also assume $a,b\ge 2$ since it is also not hard to see that:
$$
f_n(1,b)\to
\begin{cases}
0 & (n_0=4), & 2\mid b \\
b & (n_0=2), & 2\not\mid b
\end{cases}
$$
$$
f_n(a,1)\to
\begin{cases}
0 & (n_0=5), & 2\mid a \\
a & (n_0=3), & 2\not\mid a
\end{cases}
$$
- If $a=2$ then assume $b$ is odd, and if $b=2$ then assume $a$ is odd, since:
$$\begin{array}{}
f(2,2k)\to 0, & (n_0=5) \\
f(2k,2)\to 0, & (n_0=6) \\
\end{array}$$
- Assume $a,b$ are not solutions to "$0=ab\bmod(a+b)$", since then $f_n(a,b)\to0, (n_0=3)$.
Looking at the last assumption, continuing to try to characterize "short patterns" via such equations $(x_{i}\cdot x_{i+1})\bmod (x_{i}+x_{i+1})=x_{i+2}$ looks like a never ending spiral of problems.
Instead, alternative ways are needed to find and prove $n_0^{\text{max}}$ and the remainder of the claim.
This reminds me more and more of the Collatz conjecture. In other words, this recursion could be as hard as that famous unsolved conjecture.
Nonlinear recurrences are generally chaotic. Even more, recurrence depending on modulo operation does not help at all.
About the $f_n(a,b,c)$ recursive function
Trying to characterize patterns here seems too hard. Even restricting to only $f_n(1,1,c),c\in\mathbb N$ sequences, I'm not seeing any useful structures.
I have computationally examined starting conditions $(1,1,c),c\in\mathbb N$. The elements of $F$ can get large, but they seem to have a lot of small factors. Hence, I will write them down in terms of their prime factorization.
It seems $n_0$ could get arbitrarily large, so I compiled the table of record $n_0$'s for $(1,1,c)$:
$$\begin{array}{ccl}
(a,b,c) & n_0 & F \\
(1,1,1) & 1& \left(\begin{array}{}1\end{array}\right)\\
(1,1,2) & 6& \left(\begin{array}{}0\end{array}\right)\\
(1,1,3) & 8& \left(\begin{array}{}2\end{array}\right)\\
(1,1,4) & 9& \left(\begin{array}{}16\end{array}\right)\\
(1,1,5) & 10& \left(\begin{array}{}0\end{array}\right)\\
(1,1,8) & 19& \left(\begin{array}{}0\end{array}\right)\\
(1,1,9) & 143& \left(\begin{array}{}2^{33}\cdot5^2,\\2^{33}\cdot5^2,\\2^{31}\cdot5^2\cdot13,\\2^{31}\cdot5^2\cdot19\end{array}\right)\\
(1,1,18) & 493& \left(\begin{array}{}2^{73}\cdot5^3\cdot7^2\cdot11^2,\\2^{71}\cdot5^3\cdot7^3\cdot11^2,\\2^{70}\cdot5^4\cdot7^2\cdot11^2,\\2^{70}\cdot5^4\cdot7^2\cdot11^2\end{array}\right)\\
(1,1,73) & 1169& \left(\begin{array}{}2^{183}\cdot5^{13}\cdot7^{9}\cdot11^{6}\cdot13^{1}\cdot17^{2}\end{array}\right)\\
(1,1,128) & 4351& \left(\begin{array}{}2^{685}\cdot5^{83}\cdot7^{35}\cdot11^{6}\cdot13^{1}\cdot17^{2}\cdot19^{2}\cdot23^{2}\end{array}\right)\\
(1,1,877) & 5529& \left(\begin{array}{}2^{800}\cdot5^{87}\cdot7^{42}\cdot11^{13}\cdot13^{9}\cdot17^{1}\cdot19^{6}\cdot83^{1}\end{array}\right)\\
(1,1,1774) & 8637& \left(\begin{array}{}2^{1298}\cdot5^{140}\cdot7^{59}\cdot11^{20}\cdot13^{9}\cdot17^{4}\cdot23^{1}\cdot29^{2}\cdot79^{2}\cdot107^{1},\\2^{1298}\cdot5^{140}\cdot7^{59}\cdot11^{20}\cdot13^{9}\cdot17^{4}\cdot23^{1}\cdot29^{2}\cdot79^{2}\cdot107^{1},\\2^{1298}\cdot5^{140}\cdot7^{58}\cdot11^{20}\cdot13^{10}\cdot17^{4}\cdot23^{1}\cdot29^{2}\cdot79^{2}\cdot107^{1},\\2^{1298}\cdot5^{142}\cdot7^{58}\cdot11^{20}\cdot13^{9}\cdot17^{4}\cdot23^{1}\cdot29^{2}\cdot79^{2}\cdot107^{1},\\2^{1298}\cdot5^{142}\cdot7^{58}\cdot11^{20}\cdot13^{9}\cdot17^{4}\cdot23^{1}\cdot29^{2}\cdot79^{2}\cdot107^{1}\end{array}\right)\\
\dots & \dots & \dots
\end{array}$$
Another observation is that cycles seem to be able to be of arbitrary length as well. For example, $f_n(1,1,7618)$ converges to a cycle $F$ of $32$ elements (at $n_0=556$):
$$\left(\begin{array}{l}
2^{106}\cdot5^{12}\cdot7^{4}\cdot13^{1}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{2}\cdot19^{1},\\
2^{106}\cdot5^{14}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{2},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{2}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{14}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{2},\\
2^{109}\cdot5^{13}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{108}\cdot5^{12}\cdot7^{4}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{2}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1}\cdot61^{1},\\
2^{112}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{107}\cdot5^{13}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{14}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{4}\cdot13^{1}\cdot19^{1},\\
2^{108}\cdot5^{12}\cdot7^{4}\cdot13^{1}\cdot19^{1},\\
2^{109}\cdot5^{13}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{14}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{2},\\
2^{108}\cdot5^{12}\cdot7^{4}\cdot13^{1}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{4}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{110}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1},\\
2^{106}\cdot5^{12}\cdot7^{3}\cdot13^{1}\cdot19^{1}\cdot31^{1}
\end{array}\right)$$
Even if we only observe $c$'s such that $f_n(1,1,c)\to 0$, the $n_0$'s still seem to grow arbitrarily.
For example, $f_n(1,1,417)$ converges to $0$ after $n_0=448$ steps.
What is worse here compared to the "simpler" recursion, is that here the "decreasing pattern", if it exists, does not look easy.
$$ \begin{array}{} f_n(6k+0,6k-6)\to 0, & n_0=k+1, & \text{(Ends as: $...,12,6,0.$)}\ f_n(6k+2,6k-4)\to 0, & n_0=k+5, & \text{(Ends as: $...,14,8,2,6,4,4,0.$)}\ f_n(6k+4,6k-2)\to 0, & n_0=k+3, & \text{(Ends as: $...,16,10,4,12,0.$)}\ \end{array} $$
– Vepir Dec 31 '19 at 20:48