The OEIS sequence A248049 is defined by

an=(an−1+an−2)(an−2+an−3)an−4with a0=2,a1=a2=a3=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 a0=−1,a1=a2=a3=1 but now a144 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

unprovenobservations about its

factorization algebraically and p-adically for a few small values of p, but nothing that would prove integrality. For example, if

x0,x1,x2,x3 are indeterminates, and

we use initial values of

a0=x0,a1=x1,a2=x2,a3=x3 and x4:=x1+x2, with the same recursion, then an has denominator a monomial in

x0,x1,x2,x3,x4 with

exponents from OEIS sequence A023434.

Since x0=x4=2 with the original sequence I can’t prove that the numerator

has enough powers of 2 to compensate. Another

example is that a12n+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(an)≈1.25255cn

where c is the plastic constant OEIS sequence A060006. Note

that x4−x3−x2+1=(x−1)(x3−x−1) and

c is the real root of the cubic factor.Can anyone give a proof of integrality of A248049?

**Answer**

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 an sequence suggests the Ansatz

an=pnpn+1pn+2pn+3

where pn is some sequence yet to be determined.

The sequence an is supposed to satisfy a recurrence. For example, we must have

a4a0=(a1+a2)(a2+a3).

Rewriting this equation in terms of p and solving for

p7 gives the rational solution

p7=(p1+p5)(p2+p6)p3p4p0p1p6.

Rewrite this as a polynomial equation to get

p6p0p7p1=(p1+p5)p3(p2+p6)p4.

Now suppose that pn satisfies the recurrence

pn=pn−3pn−1+pn−5pn−6.

Check that this recurrence satisfies the polynomial

equation for p7.

From the pn recurrences for n=9 and

n=6 we have

p9p3=(p4+p8)p6 and p6p0=(p1+p5)p3.

Combine the two equations to simply get

(p0+p4+p8)p6=(p1+p5+p9)p3.

This implies that the number

c:=p0+p4+p8p3p4p5=p1+p5+p9p4p5p6

is constant and thus, the sequence pn satisfies the

equation

pn+pn−4+pn−8=cpn−3pn−4pn−5.

By the way, defining another constant

s:=√(a2+a1)a2a0/(a3a1)

implies the equation

c=s(a0+a1+a2)(a1+a2+a3)a0a2(a1+a2),

or more symmetrically, this can be written as

c=(a0+a1+a2)(a1+a2+a3)√a0a1a2a3(a1+a2).

Given values of p0 and p1 then

p2=s/p0 and p3=a0/(p1s)

while the two sequences are related by

pn=pn−4an−3/an−4.

If the sequence terms p0,p1,…,p7 are integers

and the constant c is an integer, then this implies that

pn is an integer sequence, and also an using the

Ansatz. In our case, c=6 and the sequence pn

begins 1,1,1,1,1,1,2,3,4,10,33,140,….

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

anan−3=a2n−1+a2n−2

and now with a constant

c:=a2n+a2n+1+a2n+2anan+1an+2

such that the sequence an also satisfies

an+an−3=can−1an−2.

**Attribution***Source : Link , Question Author : Somos , Answer Author : Somos*