Is OEIS A248049 an integer sequence?

A proof of integrality can be based on elementary algebra and some fortunate observations. The Darij Grinberg comments reminded me of some of my work I did in 2013 and which I did
not follow up on adequately.

Factorization of the $\,a_n\,$ sequence suggests the Ansatz

$$a_n = p_n p_{n+1} p_{n+2} p_{n+3}$$

where $\,p_n\,$ is some sequence yet to be determined.
The sequence $\,a_n\,$ is supposed to satisfy a recurrence. For example, we must have

$$a_4a_0 = (a_1+a_2)(a_2+a_3).$$

Rewriting this equation in terms of $\,p\,$ and solving for
$\,p_7\,$ gives the rational solution

$$p_7 = \frac{(p_1 + p_5)(p_2 +p_6) p_3 p_4}{p_0 p_1 p_6}.$$

Rewrite this as a polynomial equation to get

$$p_6p_0p_7p_1 = (p_1 + p_5)p_3(p_2 + p_6)p_4.$$

Now suppose that $\,p_n\,$ satisfies the recurrence

$$p_n = p_{n-3}\frac{p_{n-1} + p_{n-5}}{p_{n-6}}.$$

Check that this recurrence satisfies the polynomial
equation for $\,p_7.\,$

From the $\,p_n\,$ recurrences for $\,n=9\,$ and
$\,n=6\,$ we have

$$p_9p_3 = (p_4 + p_8)p_6 \quad \text{ and } \quad p_6p_0 = (p_1 + p_5)p_3.$$

Combine the two equations to simply get

$$(p_0 + p_4 + p_8)p_6 = (p_1 + p_5 + p_9)p_3.$$

This implies that the number

$$c := \frac{ p_0 + p_4 + p_8 }{p_3 p_4 p_5} = \frac{ p_1 + p_5 + p_9 }{p_4 p_5 p_6}$$

is constant and thus, the sequence $\,p_n\,$ satisfies the
equation

$$p_{n}+p_{n-4}+p_{n-8} = c\,p_{n-3}p_{n-4}p_{n-5}.$$

By the way, defining another constant

$$s := \sqrt{(a_2+a_1)a_2a_0/(a_3a_1)}$$

implies the equation

$$c = s\frac{(a_0+a_1+a_2)(a_1+a_2+a_3)}{a_0a_2(a_1+a_2)},$$

or more symmetrically, this can be written as

$$c = \frac{(a_0+a_1+a_2)(a_1+a_2+a_3)} {\sqrt{a_0a_1a_2a_3(a_1+a_2)}}.$$

Given values of $\,p_0\,$ and $\,p_1\,$ then
$\,p_2 = s/p_0\,$ and $\,p_3 = a_0/(p_1s)\,$
while the two sequences are related by
$\,p_n = p_{n-4}a_{n-3}/a_{n-4}.\,$

If the sequence terms $\,p_0, p_1,\dots, p_7\,$ are integers
and the constant $\,c\,$ is an integer, then this implies that
$\,p_n\,$ is an integer sequence, and also $\,a_n\,$ using the
Ansatz. In our case, $\,c=6\,$ and the sequence $\,p_n\,$
begins $\,1,1,1,1,1,1,2,3,4,10,33,140,\dots.\,$
This sequence was known to me in 2013 but I
do not think I connected it to A248049 at that time.

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A simpler example of a sequence similar to $\,p\,$
is OEIS A064098 with
$$a_na_{n-3} = a_{n-1}^2 + a_{n-2}^2$$
and now with a constant
$$c := \frac{a_n^2+a_{n+1}^2+a_{n+2}^2} {a_na_{n+1}a_{n+2}}$$
such that the sequence $\,a_n\,$ also satisfies
$$a_n + a_{n-3} = c\,a_{n-1}a_{n-2}.$$