# $N \equiv 3 (\textrm{mod } 4)$ and Collatz conjecture

Can the Collatz conjecture also be interpreted as behaviour, transformation of number of form $N\equiv 3(\textrm{mod }4)$ to the form of $N\equiv 1(\textrm{mod }4)$

Because integers of the form $N\equiv 3(\textrm{mod }4)$ can be later always only ones be divided by $2$. Number of form $N\equiv 1(\textrm{mod }4)$ have much more diversity.

$3,7,11,15,19,23,29,35, \ldots, N \equiv 3(\textrm{mod }4)$(distance $4$ between numbers)

using $(3n+1)/2$ or applying $+2,+4,+6,\ldots$ turns into

$5,11,17,23,29,35,41,47,\ldots, N \equiv 1 (\textrm{mod }4)$ or $3$ (50%) (distance $6$ between numbers)

again using $(3n+1)/2$ or applying $+3,+6,+9$ turns into

$8,17,26,35,44,53,62,71,\ldots$ (distance 9 between numbers), $N(\textrm{mod }4)$ is various.

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Most congruence class ideas only are predictable for so many terms.

Your class $4k+3$ is odd, so the next is $3(4k+3)+1=12k+10$, which is even, so that the next is $6k+5$. [Note here the modulus has shifted from 4 to 6, one source of why one can go only so far to predict further values.]

Now you can go another step since $6k+5$ is odd, so the next is $3(6k+5)+1=9k+8$. At this point your sequence has proceeded: $$4k+3,\ 12k+10,\ 6k+5,\ 9k+8.$$ This is where it stops, since one cannot decide whether $9k+8$ is even or odd.

To go another term you could replace $k$ everywhere by $2k$ or by $2k+1$, so that your initial number would be $8k+3$ or $8k+7$ respectively, and in this case you could go another step for either of these cases since you would know whether the above last term $9k+8$ (after replacing $k$ by $2k$ or $2k+1$) is even or odd.

When I've monkeyed around with this approach, I've come up with lots of classes, and the initial number in front of $k$ typically got large, with a lot of factors, and the classes of course did not cover all positive integers (or the problem would be easy and solved already). Still, it is entertaining to look at the congruence cases.

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see the table for the complete set of odd integers below in my answer – Gottfried Helms Dec 24 '12 at 0:57

To add to @coffemath approach: we can separate the class of odd numbers a,b in the following way such that $b=(3a+1)/2^A$ where $A$ is the value which makes b an odd integer: $$\begin{array} {r|c|l||r|c|l} & k \ge 0& & && k\gt 0\\ A & a=& b= & & A & a= & b= \\ \hline \\ 0 & 2^0 \cdot 2 k +0 & k & & 1 & 2^1 \cdot 2 k -1 & 6 k - 1 \\ 2 & 2^2 \cdot 2 k +1 & 6 k + 1 & &3 & 2^3 \cdot 2 k -3 & 6 k - 1 \\ 4 & 2^4 \cdot 2 k +5 & 6 k + 1 & &5 & 2^5 \cdot 2 k -11 & 6 k - 1 \\ 6 & 2^6 \cdot 2 k +21 & 6 k + 1 & &7 & 2^7 \cdot 2 k -43 & 6 k - 1 \\ &\cdots & \cdots & && \cdots & \cdots \\ A & 2^A \cdot 2 k + r & 6 k + 1 & &A & 2^A \cdot 2 k -s & 6 k - 1 \\ \end{array}$$ where the residues in the columns below the a are dependent on A by $$r = { 2^A -1 \over 4-1} \qquad s ={ 2^A +1 \over 4-1}$$ Of course, the entry at $A=0$ should not occur in this table, if we look only at odd a but leave it there for completeness/smoothness of the table.
(That this table covers all odd positive integers is a nice exercise, left to the reader...)

The known "trivial" cycle is that where a=1 in the category A=2 where k=0 such that $$2^2 \cdot 2 k + 1 \to 6 k + 1 \quad||_{k=0}\quad 2^2 \cdot 2 \cdot 0+1 \to 1$$
If we look also at the negative odd integers we have the cycle $-1 \to -1$ at A=1, k=0 such that $$2^1 \cdot 2 k - 1 \to 6 k - 1 \quad||_{k=0}\quad 2^1 \cdot 2 \cdot 0 -1 \to -1$$
and the cycle $-5 \to -7 \to -5$ by $A_1=1$, $k_1=-1$ and $A_2=2$, $k_2=-1$ : $$2^1 \cdot 2 k - 1 \to 6 k - 1 \quad ||_{k=-1} \quad -5 \to -7 \\ 2^2 \cdot 2 k + 1 \to 6 k + 1 \quad||_{k=-1}\quad -7 \to -5 \\$$