# 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?

• If you have $p$-integrality for all $p$, then the (global) integrality follows. Mar 19, 2020 at 3:13
• Reminds me of Somos Sequence's. Mar 19, 2020 at 11:27
• @Vepir Probably not surprising because the author of OEIS sequence A248049 is Michael Somos.
– user
Mar 19, 2020 at 11:34
• Similar to math.stackexchange.com/questions/1905063/… (note: the proof is less than 100% verified and rather unsatisfactory in its brute-force component). Mar 21, 2020 at 15:08
• @darijgrinberg Thanks for your helpful comment! The sequence $p_n$ you wrote (and several other similar sequences) was known to me early in 2013 eight years ago. A simpler recursion is $p_n p_{n-6}=(p_{n-1}+p_{n-5})p_{n-3}.$ I agree that proving integrality of this would imply integrality of A248049. Unfortunately, I did not have time then to explore all of the sequences I found and their interelations. Thanks for reminding me! Jan 28, 2021 at 3:25

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

• For me, this is almost magic. In a positive way. Jan 31, 2021 at 2:18