I found the following law and would like to know what do you think about it and if anyone can explain why this is so. Also, is this already known and proven?
Consider the following series: $$1\times2\times3, 2\times3\times4, 3\times4\times5,\ldots,n(n+1)(n+2)$$
or more compactly:
$$a_n = n(n+1)(n+2)$$
Lets take the prime factorization for every element of $a_n$:
$$2\times3, 2^3\times3, 2^2\times3\times5,2^3\times3\times5,\ldots$$
Lets take all the powers of $2$ in every prime factorization in the above series and create a sequence out of them, and then group the elements of that sequence in tuples with size $4$. The result of this we will call $p_1$. $$p_1 = (1,3,2,3),(1,4,3,4),(1,3,2,3),(1,5,4,5),\ldots$$
It seems that every $4$-tuple with odd index in $p_1$ is $(1,3,2,3)$.
Lets remove all $4$-tuples with odd positions from $p_1$ and call the result $p_2$:
$$p_2 = (1,4,3,4),(1,5,4,5),(1,4,3,4),(1,6,5,6),\ldots$$
Now $p_2$ will have $(1,4,3,4)$ at every odd position. If we take the $p_3$ sequence to have only those elements of $p_2$ that are with even indices, $p_3$ will have $(1,5,4,5)$ at every odd position. And so on... $p_n$ will have $(1, n+2, n+1,n+2)$ at every odd position.
Let $n=8k+1$. Then $a_{8k+1}=(8k+1)(8k+2)(8k+3)=2(8k+1)(4k+1)(8k+3)$, which shows that $2\mid a_{8k+1}$ but $4\not\mid a_{8k+1}$. Similarly,
$$a_{8k+2}=8(4k+1)(8k+3)(2k+1)$$ $$a_{8k+3}=4(8k+3)(2k+1)(8k+5)$$ $$a_{8k+4}=8(2k+1)(8k+5)(4k+3)$$
show that indeed $(1,3,2,3)$ always repeat itself.
The next part is showing that $a_{16k+5}$, $a_{16k+6}$, $a_{16k+7}$ and $a_{16k+8}$ have the same property. Indeed $$a_{16k+5}=2(16k+5)(8k+3)(16k+7)$$ $$a_{16k+6}=16(8k+3)(16k+7)(2k+1)$$ $$a_{16k+7}=8(16k+7)(2k+1)(16k+9)$$ $$a_{16k+8}=16(2k+1)(16k+9)(8k+5),$$ so the pattern $(1,4,3,4)$ will also repeat forever.
In general, $$\begin{align} a_{8km+4k-3}&=(8km+4k-3)(8km+4k-2)(8km+4k-1)\\&=2(8km+4k-3)(4km+2k-1)(8km+4k-1)\\ \\ a_{8km+4k-2}&=(8km+4k-2)(8km+4k-1)(8km+4k)\\ &=8k(4km+2k-1)(8km+4k-1)(2m+1)\\ \\ a_{8km+4k-1}&=(8km+4k-1)(8km+4k)(8km+4k+1)\\ &=4k(8km+4k-1)(2m+1)(8km+4k+1)\\ \\ a_{8km+4k}&=(8km+4k)(8km+4k+1)(8km+4k+2)\\&=8k(2m+1)(8km+4k+1)(4km+2k+1) \end{align}$$ generates $(1, 3+v(k), 2+ v(k), 3+v(k))$, denoting by $v(k)$ the power of $2$ in the factorization of $k$.