Property of composite numbers

Question

Let $a(n)$ be the sequence of composite numbers (starting from $4$). Let $$b(n)=a(n-1)a(n-2) \operatorname{mod} a(n)$$ Obviously, $b(1)=b(2)=0$.

I conjecture that with the only exception for the $b(3)=0$ all terms are belong to $\left\lbrace2,3,6,8\right\rbrace$.

My second conjecture is that the sequence $b(n)$ can be partitioned into blocks $\left\lbrace8\right\rbrace$ and $\left\lbrace3\underbrace{2\cdots2}_{2k-1}6\right\rbrace$.

My third conjecture is that for $b(n)=6$ the remainders of the division $a(n) \operatorname{mod} b(n)$ are belong to $\left\lbrace0,2\right\rbrace$.

My fourth conjecture is that we take the sequence $a_1(n)$ of composite numbers without Sarrus numbers instead (simply as complement of primes and Sarrus numbers), then for $b_1(n)=6$ we have violation of the rule above somewhere, exactly $a_1(n) \operatorname{mod} b_1(n) = 4$. If so, then $b_1(n)-1$ is the Sarrus numbers (since it is divisible by $3$). The sequence of such Sarrus numbers begins

561, 1905, 8481, 18705, 23001, 87249, 154101, 206601, 215265, 289941, 427233, 526593

Here similarly $$b_1(n)=a_1(n-1)a_1(n-2) \operatorname{mod} a_1(n)$$ and Sarrus numbers is a composite odd numbers $n$ such that $n$ divides $2^n - 2$.

Is there a way to prove all or part of those conjectures?

Answer 1

For any $m$ we have that either $m+1$ or $m+2$ is composite. Therefore a tuple of three consecutive composites must be of one of the following forms: $$(k-4,k-2,k),(k-3,k-2,k),(k-3,k-1,k),(k-2,k-1,k).$$ Since we have $(k-a)(k-b)\equiv ab\pmod k$, we get the four cases you ask for with $k=a(n)$. The second conjecture follows easily as well - your blocks correspond to blocks of composites between consecutive primes.

For the third conjecture, note that if $b(n)=6$, then the pattern above must be $(k-3,k-2,k)$, meaning $k-1$ is a prime, meaning $k-1\equiv 1$ or $5\pmod 6$, so $k\equiv 2$ or $0\pmod 6$.

For the fourth conjecture, I'm assuming you meant $a_1(n)-1$ is a Sarrus number. This is true - by a similar analysis as above, we get that if $b_1(n)=6$, then $a_1(n)-1$ must be either a prime or a Sarrus number. But $a_1(n)-1\equiv 3\pmod 6$ under the assumption on $a_1\pmod b_1$, so it can't be prime.