The least value of $b$ such that $2^{3^{\cdots^a}}\leq b^{(b-1)^{\cdots^2}}$

Question

I'm interested in an upper bound for the minimal positive integer $b$ for which $$2^{3^{4^{\cdots^a}}}\leq b^{(b-1)^{\cdots^{3^{2}}}}$$ holds, given a positive integer $a\geq2$.

If possible, but I bet it's not, a closed formula for $b$ would be great...

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A natural but rather unuseful first step would be to take logarithms some times:

$$3^{4^{\cdots^{a}}}\ln(2)\leq(b-1)^{\cdots^{3^2}}\ln(b)$$

But after doing this two more times one will obtain a sum in the logarithm and things get ugly.

Any estimation of $b$ would be interesting.

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Edit: Thank's to Ross Millikan's answer below, I'm somehow expecting the answer will be $b=a+1$, but until now this is not much more then intuition.

Answer 1

We have the following result:

Theorem. Let $a\geq3$. If $n\geq2$ such that $$a \geq\left((n+2)^{\cdots^2}\right)^{\exp(1+1/n)} \quad\text{resp.}\quad a \leq\left((n+2)^{\cdots^2}\right)^{\exp(-1-1/n)}$$ then $$2^{3^{4^{\ldots^a}}}\geq (a+n)^{(a+n-1)^{\ldots^{3^2}}} \quad\text{resp.}\quad 2^{3^{4^{\ldots^a}}}\leq (a+n)^{(a+n-1)^{\ldots^{3^2}}}$$

For $n\geq2$, $$\left((n+2)^{\cdots^2}\right)^{\exp(1+1/n)} \leq\left((n+3)^{\cdots^2}\right)^{\exp(-1-1/n)}$$ This means we have determined our minimal $b$ for all $a$ between two such numbers.

A lot remains unknown, but at least it gives an idea of how fast the answer grows: basically it says that the size of a power tower is determined by its tail: if we disregard the first $a-2$ levels, the towers $$2^{\cdots^a} \quad\text{and}\quad (a+n)^{\cdots^2}$$ have tails $$a \quad\text{and}\quad (n+2)^{(n+1)^{\cdots^2}}$$ If one of the tails is considerably larger (in the sense as in the theorem above), the entire tower will be larger.

More precisely:

Estimating a power tower by its tail. Let $a_1,\ldots,a_n\geq e$ be positive real numbers. Let $2\leq m\leq n+1$ be an integer. Then $$\left|\log_m{a_1}^{\ldots^{a_n}}-\log\log{a_{m-1}}^{\ldots^{a_n}}\right|\leq\sum_{k=2}^{m-1}\frac{\log\log a_{k-1}}{\log{a_k}^{\ldots^{a_n}}}$$ where $\log_m=\log\circ\cdots\circ\log$ ($m$ times) and any ill-defined term equals $0$ by convention.

The proof is nothing but repeatedly taking logarithms and using $|\log x-\log y|\leq\frac{|x-y|}{\min(x,y)}$ (Mean Value Theorem); in particular $\leq|x-y|$ for $x,y\geq1$. The hardest part was formulating it ;-)

Proof. By induction on $m$ for fixed $n$. It is clear for $m=2$. If true for some $m\leq n$, we have $$\left|\log_{m+1}{a_1}^{\ldots^{a_n}}-\log\left(\log{a_m}^{\ldots^{a_n}}+\log\log a_{m-1}\right)\right| \leq \sum_{k=2}^{m-1}\frac{\log\log a_{k-1}}{\log{a_k}^{\ldots^{a_n}}}$$ and we conclude using $$\left|\log\left(\log{a_m}^{\ldots^{a_n}}+\log\log a_{m-1}\right)-\log\log{a_m}^{\ldots^{a_n}}\right| \leq \frac{\log\log a_{m-1}}{\log {a_m}^{\ldots^{a_n}}}$$ $\square$

This can certainly be improved, especially as we used the weak $|\log x-\log y|\leq|x-y|$ and only then the stronger $|\log x-\log y|\leq\frac{|x-y|}{\min(x,y)}$. But this is a nice way to present it while maintaining a good upper bound. We want to apply this with $a_1=2<e$, so let's look if this is a problem for the proof. The first inequality stays valid, because for $m=2$ it says nothing. In the second we have to add $\log\log2$ to the denominator and take the absolute value of the RHS.

Proof of the Theorem.

The estimates are crude, because precision is not our main interest for now. Let $a\geq3$ and $n\geq2$. We have $$\begin{align*}\left|\log_a2^{\cdots^a}-\log\log a\right| &\leq\frac{-\log\log2}{\log3+\log\log2}+\sum_{k=3}^{a-1}\frac{\log\log k}{\log(k+1)^{\cdots^a}}\\ &\leq0.51+a\cdot\frac{\log\log a}{a^2}+\frac{\log\log(a-2)}{a\log(a-1)}+\frac{\log\log(a-1)}{\log a} \end{align*}$$ as $(k+1)^{\cdots^a}\geq3^{a^2}$ for $3\leq k\leq a-3$. (negative terms are $0$ by convention) This is $\leq1$ for all $a$ (W/A).

For the other tower, $$\begin{align*}\left|\log_a(a+n)^{\cdots^2}-\log\log (n+2)^{\cdots^2}\right| &\leq\sum_{k=2}^{a-1}\frac{\log\log(a+n+2-k)}{\log(a+n+1-k)^{\cdots^2}}\\ &=\sum_{k=1}^{a-2}\frac{\log\log(n+2+k)}{\log(n+1+k)^{\cdots^2}}\\ &\leq\sum_{k=1}^\infty\frac{\log\log(n+2+k)}{(n+k)(n+k-1)\log(n+1+k)}\\ &\leq0.37\sum_{k=0}^\infty\frac1{(n+k)(n+k-1)}\\ &=\frac{0.37}{n-1}<1/n \end{align*}$$

Thus if $$\log\log a\geq\log\log(n+2)^{\cdots^2}+1+1/n\iff a \geq\left((n+2)^{\cdots^2}\right)^{\exp(1+1/n)}$$ then $$2^{3^{4^{\ldots^a}}}\geq (a+n)^{(a+n-1)^{\ldots^{3^2}}}$$

and if $$\log\log a\leq\log\log(n+2)^{\cdots^2}-1-1/n\iff a \leq\left((n+2)^{\cdots^2}\right)^{\exp(-1-1/n)}$$ then $$2^{3^{4^{\ldots^a}}}\leq (a+n)^{(a+n-1)^{\ldots^{3^2}}}$$

Intuition about power towers

Higher towers are generally larger. Towers whose highest levels are heavy are easier to estimate (as we see from the main estimate). High and light levels aren't really levels (they can be grouped as one).

Answer 2

If you want an estimate, mine would be $b=a+1$ The most important thing about power towers is the number of levels-nothing can compete with that. For the same number of levels ($a=b$), your stacking on the left is clearly greater, so $b \gt a$. Your idea of taking logs is a good one. Then if you recognize that with those towers around, the difference between $\log 2$ and $\log b$ is trivial, you can cancel them and take another log. Keep going. Clearly this is a huge handwave, but one I believe in.