For this inequality $$3^i\cdot\left\{\frac{2^i\cdot3^{2^{\lceil i\cdot \log_23\rceil-i-2}}}{2^{\lceil i\cdot \log_23\rceil}3^i}\right\}-\frac{2^i}{2^{\lceil i\cdot \log_23\rceil}}\leqslant 2^{\lceil i\cdot \log_23\rceil}\cdot\left\{\frac{2^i\cdot3^{2^{\lceil i\cdot \log_23\rceil-i-2}}}{2^{\lceil i\cdot \log_23\rceil}3^i}\right\}-1$$

where $\{x\}$ is the fractional part of $x$, and $i>6$ (to have integers on both sides),

Show that, for any $i$, this cannot be equal.

Note that $3^i\{x\}<2^{\lceil i\cdot \log_23\rceil}\{x\}$ (the fractional part is the same on both sides). Also note that for the exponents in the fractional part when $i>6$, we have $i<2^{\lceil i\cdot \log_23\rceil-i-2}$ and obviously $i<{\lceil i\cdot \log_23\rceil}$


Any positive odd integer can be written as $$n_0=a\cdot 2^i-1$$

It is well known that in a Collatz sequence, $n_0$ goes straight to $a\cdot 3^i-1$ in exactly $i$ steps of the combined Collatz function $T(n)=\frac{3n+1}{2}$

Now this number being even, it needs to be shaved of a few factor $2$.

I am looking at $n_j=\frac{a\cdot 3^i-1}{2^{\lceil i\cdot \log_2\frac{3}{2}\rceil}}$, the first $n_j$ for which $\frac{3^i}{2^j}=\frac{3^i}{2^{\lceil i\cdot \log_23\rceil}}<1$ ($n_j$ can be odd or even)

Now in this particular scenario, it can be shown that $$n_j \le n_0$$ which is the simplified version of the equation on top, but I want to show that $$n_j < n_0$$

e.g. for $i=7$: $n_0=3\cdot 2^7-1=383$ Goes up to $6560=3\cdot 3^7-1$ and is shaved down to $n_j=205$ $$\{383,575,863,1295,1943,2915,4373,6560,3280,1640,820,410,205\}$$ Reminder: $3^i=2^{i\cdot \log_23}$

EDIT: Another way would be to show that $n_j$ or $n_0\neq\frac{3^i-2^i}{2^{\lceil i\cdot \log_23\rceil}-3^i}$

Here is a list of $n_0$ values for each $i>=1$ (they can be found using formula above for $i>6$): $$\{1,3,23,15,95,575,383,255,5631,...\}$$ and as said in a comment they are found every $2^{\lceil i\cdot \log_23\rceil}$ which means that numbers concerned by this post are: $$\{4k+1, 16k+3, 32k+23, 128k+15, 256k+95, 1024k+575, 4096k+383... \}$$



  • 2
    $\begingroup$ If I'm reading this right, this would lead a proof of Collatz by induction? So I doubt that making it strict is possible with current tools, as I'm almost certain someone will have looked at this. $\endgroup$ – It'sNotALie. Aug 29 '19 at 21:32
  • $\begingroup$ I don't think it would be usefull for an induction proof, this is only 1 particular case of Collatz trajectories. All integers can be written as $ n_0=a\cdot 2^i-1$, but only one of them can be divided by 2 down to that limit (one every $2^{\lceil i\cdot \log_23\rceil}$ in fact) $\endgroup$ – Collag3n Aug 30 '19 at 5:08
  • $\begingroup$ Sketch: any even integer goes down in the first step to a smaller number for which it works. Any positive odd integer then can follow this prcedure once to get from $n_0$ to $n_j$, which is strictly smaller than $n_0$ by this strict ineq. $\endgroup$ – It'sNotALie. Aug 30 '19 at 10:05
  • $\begingroup$ Even the less strict inequality $n_j \le n_0$, easy to prove for this particular inverse v-shape scenario, has never been proven for other cases (where $\frac{3^i}{2^{\lceil i\cdot \log_23\rceil}}<1)$ $\endgroup$ – Collag3n Aug 30 '19 at 10:29
  • $\begingroup$ I added some examples in my post $\endgroup$ – Collag3n Aug 30 '19 at 11:07

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