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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?

Somos
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    If you have $p$-integrality for all $p$, then the (global) integrality follows. – WhatsUp Mar 19 '20 at 03:13
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    Reminds me of [Somos Sequence's](https://mathworld.wolfram.com/SomosSequence.html). – Vepir Mar 19 '20 at 11:27
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    @Vepir Probably not surprising because the author of OEIS sequence A248049 is Michael Somos. – user Mar 19 '20 at 11:34
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    Similar to https://math.stackexchange.com/questions/1905063/is-a276175-integer-only (note: the proof is less than 100% verified and rather unsatisfactory in its brute-force component). – darij grinberg Mar 21 '20 at 15:08
  • I don't know if this is useful, but defining $b_0=2$ and $b_n=a_{n-1}/b_{n-1}$, we obtain the slightly simpler recurrence $b_n= b_{n-2}(b_{n-1}\!+\!b_{n-3})/b_{n-4}$. This sequence also seems to be integral as far as I could check, and satisfies the additional recurrence $b_n = (b_{n-1}\!+\!b_{n-3})(b_{n-3}\!+\!b_{n-5})/b_{n-6}$. – pregunton Jan 13 '21 at 13:26
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    @pregunton Your $\,b_n\,$ is [OEIS sequence A078918](https://oeis.org/A078918). Your recurrence for $\,b_n\,$ is the first formula in the sequence entry. I seem to not indicate its relation to A248049 and also a few other sequence relations. – Somos Jan 13 '21 at 16:24
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    It seems that there is an integer sequence $\left(p_0, p_1, p_2, p_3, \ldots\right) =$ $(2, 3, 4, 10, 33, 140, 715, 7909, 277165, 19819657, 4750144681,$ $9404100347731, 260679781038269672, \ldots)$ such that every $n \geq 4$ satisfies $a_n = p_n p_{n-1} p_{n-2} p_{n-3}$. This sequence will then satisfy the recursion $p_n p_{n-1} p_{n-6} p_{n-7} = p_{n-3} p_{n-4} \left(p_{n-1} + p_{n-5}\right) \left(p_{n-2} + p_{n-6}\right)$. Maybe its integrality is easier to prove? – darij grinberg Jan 28 '21 at 03:01
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    @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! – Somos Jan 28 '21 at 03:25
  • Just came here to post that simpler recursion! (Also, I forgot to start the sequence $\left(p_0, p_1, p_2, p_3, \ldots\right)$ with three $1$s before the $2,3,4,10,33,\ldots$ part. Sorry!) – darij grinberg Jan 28 '21 at 03:39
  • Actually, we can start the sequence $\left(p_0, p_1, p_2, p_3, \ldots\right)$ with six $1$s as starting values, and it just shifts by three positions. Thus, proving the Laurent phenomenon for its $p_n p_{n-6}=(p_{n-1}+p_{n-5})p_{n-3}$ recursion would finish the proof of integrality for $p_n$ and therefore also for $a_n$. – darij grinberg Jan 29 '21 at 13:45
  • Oh, and I can show that the numbers $q_n := p_n p_{n+2}$ are integers. Indeed, they satisfy the recurrence $q_n q_{n-4} = q_{n-1} q_{n-2} + q_{n-2} q_{n-3}$ with starting values $1, 1, 1, 1$; but this is well-known to be an integral recurrence (particular case of Exercise 8.1.9 in http://www.cip.ifi.lmu.de/~grinberg/t/20f/mps.pdf ). – darij grinberg Jan 29 '21 at 13:55
  • OK, I now think I can prove everything claimed here. Will take me a while to write it up, thouhg. – darij grinberg Jan 29 '21 at 14:17
  • Actually, the detour through the $q_n$ can be avoided: Just show that each $n \geq 8$ satisfies $p_n = b p_{n-3} p_{n-4} p_{n-5} - p_{n-4} - p_{n-8}$, where $b = 6$. (The $b$ will be different for other starting values.) – darij grinberg Jan 29 '21 at 14:47
  • @darijgrinberg Yes, just show that $(p_n+p_{n-4}+p_{n-8})/(p_{n-3}p_{n-4}p_{n-5})$ is constant. – Somos Jan 29 '21 at 15:37

1 Answers1

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

Somos
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