User:Robkahn131/Sandbox

Failed Proof
that for some $n \in N$, there does NOT exist any $i \in \N$ such that the sequence $S_i = 1$.

From Definition:Ordering on Natural Numbers, we know that this hypothetical sequence $\sequence {S_k}$ must have a smallest element.

Suppose $a = \sequence {S_k}$ is the minimum value of this hypothetical sequence

We know that $a$ can not be even because


 * which contradicts our assertion that $a$ was the smallest element in the sequence.

We also know that $a \not \equiv 1\pmod 4$ because


 * and since $\sequence {S_{k+3}} < \sequence {S_k}$, this contradicts our assertion that $a$ was the smallest element in the sequence.

When $a \equiv 3\pmod 4$, we obtain


 * but we can go no further because the parity of $\sequence {S_{k+4}}$ depends on r. The results here are inconclusive.

We now have that if $a = \sequence {S_k}$ is the minimum value of our hypothetical sequence, then $a$ must reside in the residue class of $3$ (modulo $4$).

Let us now see if we can improve our filter for $a$ by removing additional residue classes from contention.

Inspecting the set of residue classes modulo $16$, we focus our attention on $3$ (modulo $16$),  $7$ (modulo $16$),  $11$ (modulo $16$) and  $15$ (modulo $16$).

We know that $a \not \equiv 3\pmod {16}$ because


 * and since $\sequence {S_{k+6}} < \sequence {S_k}$, this contradicts our assertion that $a$ was the smallest element in the sequence.

Just like $3$ (modulo $4$) shown above, the results for  $7$ (modulo $16$),  $11$ (modulo $16$) and  $15$ (modulo $16$) are inconclusive.

We now have the following:

If a counterexample to the Collatz conjecture exists, then such a counterexample would require a sequence with a smallest element.

This smallest element could only reside in the following residue classes:

(modulo $4$): $3$

(modulo $16$): $7$, $11$, $15$

(modulo $64$): $7$, $15$, $27$, $31$, $39$, $47$, $59$, $63$

(modulo $256$): $27$, $31$, $47$, $63$, $71$, $91$, $103$, $111$, $127$, $155$, $159$, $167$, $191$, $207$, $223$, $231$, $239$, $251$, $255$

As we can see, (modulo $4^n$), there are $\approx 2^{1.5\paren{n - 1} }$ residue classes where $a$ could reside which means we have a Cantor set-esque filter which allows our hypothetical minimum element $a$ to reside in an ever dwindling proportion of the possible  residue classes. (i.e. $\dfrac{2^{1.5\paren{n - 1} } } {4^n}$).

Just like the Cantor set and just like prime numbers, at increasing values of $n$, we get an ever increasing set of something (what remains of the Cantor set, prime numbers and in this case, candidate residue classes where $a$ could reside), but the proportion of the item of interest relative to all possible values heads towards zero.

And we've proved nothing.

Math is hard.