# Showing $3^iq-2^{i-p}\neq2^pq-1$, with $p:=\lceil i\,\log_23\rceil$, $q:=\left\{\frac{2^{i}\,3^{2^{p-i-2}}}{2^p3^i}\right\}$, $i>6$

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 $$i<{\lceil i\cdot \log_23\rceil}$$

$$Background$$

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: $$\{4k+1, 16k+3, 32k+23, 128k+15, 256k+95, 1024k+575, 4096k+383... \}$$

PARI/GP

nj(i)=3^i*frac((2^i*3^(2^(ceil(i*log(3)/log(2))-i-2)))
/(2^ceil(i*log(3)/log(2))*3^i))-2^i/2^ceil(i*log(3)/log(2))

n0(i)=2^ceil(i*log(3)/log(2))*
frac((2^i*3^(2^(ceil(i*log(3)/log(2))-i-2)))
/(2^ceil(i*log(3)/log(2))*3^i))-1

• 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. – It'sNotALie. Aug 29 '19 at 21:32
• 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) – Collag3n Aug 30 '19 at 5:08
• 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. – It'sNotALie. Aug 30 '19 at 10:05
• 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 $\frac{3^i}{2^{\lceil i\cdot \log_23\rceil}}<1)$ – Collag3n Aug 30 '19 at 10:29
• I added some examples in my post – Collag3n Aug 30 '19 at 11:07