Is OEIS A248049 an integer sequence? The OEIS sequence A248049 is defined by
$$ a_n \!=\! \frac{(a_{n-1}\!+\!a_{n-2})(a_{n-2}\!+\!a_{n-3})}{a_{n-4}} \;\text{with }\; a_0\!=\!2, a_1\!=\!a_2\!=\!a_3\!=\!1.$$
is apparently an integer sequence but I have no proofs. I have numerical evidence using PARI/GP and Mathematica only. It is a real
problem because its companion
OEIS sequence A248048 has the same
recursion with $\,a_0=-1, a_1=a_2=a_3=1\,$ but now $\,a_{144}\,$ has a denominator of $2$. There is a
resemblance to the Somos-4 sequence but that probably won't help with an integrality proof.
I have some interesting unproven observations about its
factorization algebraically and $p$-adically for a few small values of $p$, but nothing that would prove integrality. For example, if
$\,x_0,x_1,x_2,x_3\,$ are indeterminates, and
we use initial values of
$$ a_0=x_0,\; a_1=x_1,\; a_2=x_2,\; a_3=x_3 \;\text{ and }\;
x_4 := x_1+x_2,$$ with the same recursion, then $\,a_n\,$ has denominator a monomial in
$\,x_0,x_1,x_2,x_3,x_4\,$ with
exponents from OEIS sequence A023434.
Since $\,x_0=x_4=2\,$ with the original sequence I can't prove that the numerator
has enough powers of $2$ to compensate. Another
example is that $\,a_{12n+k}\,$ is odd for $\,k=1,2,3\,$ and
even for the other residue classes modulo $12$. I also have
some further observations about its $2$-adic valuation
behavior which I can't prove.
By the way, the sequence grows very fast. My best
estimate is $\,\log(a_n) \approx 1.25255\, c^n\,$
where $\,c\,$ is the plastic constant OEIS sequence A060006. Note
that $$x^4-x^3-x^2+1 = (x-1)(x^3-x-1) $$ and
$\,c\,$ is the real root of the cubic factor.
Can anyone give a proof of integrality of A248049?
 A: 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.

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
$$ a_n + a_{n-3} = c\,a_{n-1}a_{n-2}. $$
