Is OEIS A248049 an integer sequence?

The OEIS sequence A248049 is defined by

an=(an1+an2)(an2+an3)an4with 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 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
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 x4x3x2+1=(x1)(x3x1) and
c is the real root of the cubic factor.

Can anyone give a proof of integrality of A248049?


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


where pn is some sequence yet to be determined.
The sequence an is supposed to satisfy a recurrence. For example, we must have


Rewriting this equation in terms of p and solving for
p7 gives the rational solution


Rewrite this as a polynomial equation to get


Now suppose that pn satisfies the recurrence


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


This implies that the number


is constant and thus, the sequence pn satisfies the


By the way, defining another constant


implies the equation


or more symmetrically, this can be written as


Given values of p0 and p1 then
p2=s/p0 and p3=a0/(p1s)
while the two sequences are related by

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
and now with a constant
such that the sequence an also satisfies

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

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