How to Find the Slowest-Growing Integer Sequence with Prime Factor Constraints

Hello! This is a fascinating problem in sequence analysis and asymptotic growth theory. Let me help you think through this systematically.

What We're Solving:

You're looking for the slowest-growing integer sequence $(e_k)$ that satisfies:

• $e_1 \geq 1$
• $p^{e_k} > b^{e_{k-1}}$ for all $k \geq 2$
• Has a "closed-form" (non-recurrent) definition

Where $p$ is a prime factor of $b$, and $b$ has at least two distinct prime factors.

The Approach:

The key insight is to transform the exponential inequality into a linear one using logarithms, then find the "minimal" sequence that satisfies this constraint. We want to balance being as small as possible while still meeting the growth requirements.

Step-by-Step Analysis:

Step 1: Transform the constraint Starting with $p^{e_k} > b^{e_{k-1}}$, taking $\log_p$ of both sides: $$e_k > e_{k-1} \log_p b$$

This is crucial because it linearizes our exponential constraint!

Step 2: Understand why $\log_p b$ is irrational Since $b$ has at least two distinct prime factors, we can write $b = p^a \cdot m$ where $\gcd(p,m) = 1$ and $m > 1$. Then: $$\log_p b = a + \log_p m$$

Since $m$ contains prime factors other than $p$, $\log_p m$ is irrational, making $\log_p b$ irrational.

Step 3: Analyze your proposed solution Your sequence $e_k = r^{k-1}$ where $r = \lceil \log_p b \rceil$ satisfies the constraints, but let's check if it's truly minimal.

Step 4: The critical question - Can we do better? Here's where we need to think carefully. Your sequence grows exponentially, but the constraint only requires: $$e_k > e_{k-1} \cdot \log_p b > e_{k-2} \cdot (\log_p b)^2 > \ldots > e_1 \cdot (\log_p b)^{k-1}$$

Since $e_1 \geq 1$, we need $e_k > (\log_p b)^{k-1}$.

Step 5: The key insight Since $e_k$ must be integers and $\log_p b$ is irrational, the minimal choice is: $$e_k = \lceil (\log_p b)^{k-1} \rceil$$

But wait! This still depends on previous terms implicitly. For a truly closed form that's guaranteed to work, your choice of $e_k = (\lceil \log_p b \rceil)^{k-1}$ is actually quite clever!

The Answer:

Your proposed sequence $e_k = r^{k-1}$ where $r = \lceil \log_p b \rceil$ is likely optimal among closed-form solutions. Here's why:

• 1. Lower bound: Any valid sequence must satisfy $e_k \geq \lceil (\log_p b)^{k-1} \rceil$
• 2. Your bound: $(\lceil \log_p b \rceil)^{k-1} \geq (\log_p b)^{k-1}$, so it works
• 3. Minimality: Since $\lceil \log_p b \rceil$ is the smallest integer greater than $\log_p b$, using any smaller base would eventually fail

Rigorous justification: Any closed-form sequence that grows slower than exponential with base $\lceil \log_p b \rceil$ would eventually be dominated by $(\log_p b)^{k-1}$ for large $k$, violating the integer constraint.

Memory Tip:

Think of this as a "bootstrapping" problem - each term needs to be big enough to make the next term work, creating inevitable exponential growth. The ceiling function is your "safety margin" that ensures integer solutions exist!

Great mathematical thinking on this problem! The interplay between the irrationality of $\log_p b$ and the integer constraint is what forces the exponential growth pattern.