Does the following hold as a conjecture for maximum gaps between prime numbers? and can it be proved?

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Even though I used matrix related mathjax on the backend, the frontend is intended to be just a regular table.

$$\begin{matrix} a&X&X:explanation \\1&1 \\2&3 \\3&5 \\4&(6)&(X_2⋅2)=(3⋅2) \\5&7 \\6&(9),(10)&(X_2⋅3)=(3⋅3),(X_3⋅2)=(5⋅2) \\7&11 \\8&(12)&(X_2⋅2⋅2)=(3⋅2⋅2) \\9&(15)&(X_3⋅3)=(5⋅3) \\10&(14),(15)&(X_5⋅2)=(7⋅2),(X_2⋅5)=(3⋅5) \\11&13 \\12&(18),(20)&(X_2⋅6)=(3⋅6),(X_3⋅4)=(5⋅4) \\13&17 \\14&(21),(22)&(X_2⋅7)=(3⋅7),(X_7⋅2)=(11⋅2) \\15&(21),(25)&(X_5⋅3)=(7⋅3),(X_3⋅5)=(5⋅5) \\16&(24)&(X_8⋅2)=(12⋅2),(X_4⋅4)=(6⋅4) \\17&19 \\18&(27),(30)&(X_2⋅9)=(3⋅9),(X_3⋅6)=(5⋅6) \\19&23 \\20&(30),(28)&(X_2⋅10)=(3⋅10),(X_4⋅5)=(6⋅5),(X_5⋅4)=(7⋅4) \\.&.&. \\.&.&. \\.&.&. \end{matrix}$$

*(The parenthesis are meant to distinguish composite numbers).

The idea is:

$a$ is just the sequence of positive integers.

For $X_a$: Starting with $X_1=1$ (as an exception), the following $X_a$ has the value of the nearest odd positive integer that is prime and wasn't mentioned earlier, unless $a$ is a composite.

That makes $X_2=3$, $X_3=5$.

If $a$ is a composite, $X_a$ is composed based on the factors of $a$, which results in a set that can have multiple distinct values.

$$ d(a)=\{m∈N:m | a\} $$

$d(a)$ is the set of all factors $m$ that divide $a$.

$$X_a=\{\forall m\in d(a): \frac{a}{m}⋅X_m\}$$.

(the $m$ in $X_m$ is a value for the arbitrary $a$ in $X_a$ such that for that specific value $X_{a=m}$).

For example:

The factors of $a=4$ are $2$ (besides $1$), $X_2=3$, and finally $2⋅X_2=2⋅3=6$.

Another Example:

The factors of $a=6$ are $2$ and $3$ (besides $1$), $X_2=3$ and $X_3=5$, and finally $X_6=X_3⋅2=5⋅2=10$ and $X_6=X_2⋅3=3⋅3=9$.

Observation:

Reminder: When $a$ is a prime number, $X_a$ will also be a prime.

This $X_a$ prime number will have a value that is: $\max(X_{a+1})>X_a > a$.

$\max(X_{a+1})$ means the highest distinct value for $(X_{a+1})$.

Examples:

$\max(X_{2+1})> X_2 >2 = \max(5)>3>2 = 5>3>2$.

$\max(X_{3+1})>X_3 >3 = \max(6)>5>3 = 6>5>3$.

$\max(X_{5+1})> X_5 >5 = \max(9,10)>7>5=10>7>5$.

The reason why I am encouraged by this observation has to do with the fact that:

If $a$ is a composite and the nearest smaller prime to it will be $p$, then $a$ can appear as $X_y=a$ when $y<p$.

(the $y$ in $X_y$ is a replacement value for the arbitrary $a$ in $X_a$ where in that specific value$X_{a=y}$).

Example: $a=9$, smaller nearest prime $p=7$, $X_y=X_6=9$, $y=6<p=7$.

Which raises another potential conjecture that can be related:

Let $p_i$ be the set of prime numbers greater than $3$ and let $i$ be the index numbers. Between each $X_{p_{i}}$ and $X_{p_{i-1}}$, there will be an $X_c$ that has a greater value than $p_i$

($i$,$i-1$, and $c$, are values for the the arbitrary $a$ in $X_a$)

Example:

Let $p_2$ be the 2nd element in the set of prime numbers greater or equal to $3$, which is equal to $5$. that means that $p_{2-1}=3$.

Between $X_{p_{2}}=X_5=7$ and $X_{p_{1}}=X_3=5$, you have an $X_c=X_4 =6$, and $ X_c=X_4=6>p_2=5$

Does the above observation hold as a conjecture? and can it be proved?

*Any answers in regard to the second potential conjecture are also appreciated.