Show that $2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$

Is there a way to show that (with integer $n>2$) $$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$ I tried to figure a way with derivative or by looking how both side are growing, but I have some trouble with the ceiling function.

Solution 1:

With $n+k= {\lceil n \log_23\rceil}<n \log_23+1$ we can use the Rhin bound (page 160: $|µ_1\log 2+µ_2\log 3|\geq H^{-13.3}$ with $H=max(|µ_1|,|µ_2|)$)

$$|(n+k) \log2 - n \log3|>\frac 1{(n+k)^{13.3}}>\frac1{(n \log_23+1)^{13.3}}>\frac 1{n^a}$$ If we choose $a=15$ the above is true for $n>41$

Now (growth of $x^a$ vs $a^x$), $$\frac 1{n^a}>\frac{1}{(\frac{3}{2})^n}$$ is true for $n>\frac{-aW_{-1}(\frac{-\log\frac{3}{2}}{a})}{log(\frac{3}{2})}$ or $n>196$ with chosen $a=15$ ($W$ is the productlog), and we also have (using $x>log(1+x)$) $$\frac{1}{(\frac{3}{2})^n}>\log(1+\frac{1}{(\frac{3}{2})^n})=\log(1+\frac{2^n}{3^n})$$ So for $n>196$ we have $$|(n+k)\log2 - n \log3|=\log(\frac{2^{n+k}}{3^n})>\log(1+\frac{2^n}{3^n})$$ or $$\frac{2^{n+k}}{3^n}>1+\frac{2^n}{3^n}$$ $$2^{\lceil n \log_23\rceil}-3^n>2^n$$

Similarly we have $$|n \log3-(n+k-1) \log2|>\frac 1{(n+k-1)^{13.3}}>\frac1{(n \log_23)^{13.3}}>\frac 1{n^a}$$ which also holds for $a=15$, and additionally we have: $$\frac{1}{(\frac{3}{2})^n}=\frac{1}{2^{n\log_2\frac{3}{2}}}>\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}>\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$ So for $n>196$ we have $$|n \log3-(n+k-1) \log2|=\log(\frac{2\cdot 3^n}{2^{n+k}})>\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$ or $$\frac{2\cdot 3^n}{2^{n+k}}>1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}$$ $$2\cdot 3^n>2^{n+k}+2^n$$

$$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$

This leads to ($n>196$):

$$\begin{array}{|c|}\hline 2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n\\\hline\end{array}$$ and with manual checking, the left inequality holds except for $n$ in $\{1,3,5\}$ and the right inequality holds except for $n$ in $\{1,2\}$

Solution 2:

If you calculate $n log_2 (3)$ for n = 1, 2, 3, ... and look at the digits after the decimal point, you will get a rather unpredictable looking sequence of numbers from 0 to 1. Using the ceil function will increase the argument by an amount between 0 and 1, so two raised to this power will be somewhere between $3^n$ and $2 \cdot 3^n$.

Your inequality will be wrong if $n log_2(3)$ is just a tiny bit larger than an exact integer. You can calculate how close it has to be to make the inequality wrong. You can calculate how likely it would be, if those last digits were random. Since these digits are not random, but in fact predictable, it could be possible to prove this either way. Others have commented that it seems to be an unsolved problem.

$2^n$ gets small compared to $3^n$ very quickly, so I’d be willing to make a small bet that the inequality is indeed true, but would be absolutely unable to prove it.

Solution 3:

Start with $$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$

Bring things to the right side: $$2^{\lceil n \log_23\rceil}<2\cdot 3^n \cdot (1-2^n / (2\cdot 3^n))$$

Take base-2 logarithm: $${\lceil n \log_23\rceil}<1 + n \log_23 + \log_2 ( (1-2^n / (2\cdot 3^n)))$$

Since $n \log_23$ is irrational for n >= 1: $${\lfloor n \log_23\rfloor + 1}<1 + n \log_23 + \log_2 ( (1-2^n / (2\cdot 3^n)))$$

$$n \log_23 - {\lfloor n \log_23\rfloor}< - \log_2 ( (1-2^n / (2\cdot 3^n)))$$

The right hand side is about $$2^n / (2\cdot 3^n) / \ln 2$$ or $${(2/3)}^n / (2\cdot \ln 2)$$

The original inequality is true as long as $n \log_23$ is more than ${(2/3)}^n / (2\cdot \ln 2)$ above the next lower integer. This series shrinks very fast, so it’s most likely true for all n.

Using Euclid’s algorithm as if you were calculating $\gcd(\log_23, 1)$ you can find consecutive n where the distance is as small as possible, and using some software that can perform these calculations in very high precision you should be able to prove this for very large n, say $n < 10^{1,000,000}$. Which doesn’t prove it’s true for all n.