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

Question

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

EDIT(27/01/22) this can also be written this way (no limitation on $i$): $$\begin{array}{|c|}\hline (3^i\cdot(3^{-i}\mod 2^{\lceil i\cdot \log_2\frac{3}{2}\rceil})-1)\cdot 2^{-\lceil i\cdot \log_2\frac{3}{2}\rceil}\leqslant 2^i\cdot(3^{-i}\mod 2^{\lceil i\cdot \log_2\frac{3}{2}\rceil})-1\\\hline\end{array}$$

$Background$

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(ilog(3)/log(2))
         frac((2^i3^(2^(ceil(ilog(3)/log(2))-i-2)))
          /(2^ceil(ilog(3)/log(2))3^i))-1

---

Proposals to improve the PARI/GP-routines (thanks to Gottfried)

    \\ define global constants instead of calling recomputations of costly functions
    beta=log(3)/log(2)    \\ of course with sufficient internal precision, I use 200 by default
\\ then your functions become


{ nj(i)=3^ifrac((2^i3^(2^(ceil(ibeta)-i-2)))
          /(2^ceil(ibeta)3^i))-2^i/2^ceil(ibeta)   }
{ n0(i)=2^ceil(ibeta)
             frac((2^i3^(2^(ceil(ibeta)-i-2)))
              /(2^ceil(ibeta)3^i))-1    }

I think it is a good habit, to use (and reserve) one-letter variable like i,j,k always for indices of loops, vectors etc. I take $N$ for the (N)umber-of-odd-steps

   { nj(N)=3^N*frac((2^N*3^(2^(ceil(N*beta)-N-2)))
          /(2^ceil(N*beta)*3^N))-2^N/2^ceil(N*beta)   }
{ n0(N)=2^ceil(Nbeta)
             frac((2^N3^(2^(ceil(Nbeta)-N-2)))
              /(2^ceil(Nbeta)3^N))-1   }

Now a new constant internal to your function which captures the repeated "ceil()"-expression. I use "S" for it, the number of even steps, or (S)um of exponents of 2. This is depending on the argument of your function and must be computed inside your function as a local constant:

    { nj(N)=my(S=ceil(N*beta));
          3^N*frac((2^N*3^(2^(S-N-2)))/(2^S*3^N))-2^N/2^S   }
{ n0(N)=my(S=ceil(N*beta));
      2^S*frac((2^N*3^(2^(S-N-2)))/(2^S*3^N))-1    }

Next, another local constant with letter $B$, because I've to use often the difference $S-N=B$ and cancel $2^S/2^N = 2^B$

    { nj(N)=my(S=ceil(N*beta),B=S-N);
          3^N    * frac(3^(2^(B-2)-N) /2^B )  -  1/2^B  }
{ n0(N)=my(S=ceil(N*beta),B=S-N);
      2^S    * frac(3^(2^(B-2)-N) /2^B )   - 1  }

Finally two more improvements are possible. One is the avoiding of intermediate gigantic numbers like $3^{2^B}$ by putting cancellations into the exponents, and second, the common expression of the frac() into an own function - to make sure the reader sees immediately, that it is the same term in both functions:

    beta = log(3)/log(2)
    fr(N,B)=  frac(2^(  beta*(2^(B-2)-N)  - B  ))
nj(N)  =  my(S=ceil(N*beta),B=S-N);    3^N * fr(N,B) -  1/2^B

n0(N)  =  my(S=ceil(N*beta),B=S-N);    2^S * fr(N,B) -  1

.

EDIT As we discussed in a linked post, this can also be written this way $$(3^N(3^{-N} \% 2^B)-1)2^{-B}\leq 2^N(3^{-N} \% 2^B)-1$$