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The OEIS sequence A248049 defined by

$$ a_n \!=\! (a_{n-1}\!+\!a_{n-2})(a_{n-2}\!+\!a_{n-3})/a_{n-4} \;\text{ with }\; a_0\!=\!2, a_1\!=\!a_2\!=\!a_3\!=\!1.$$

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?

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    $\begingroup$ If you have $p$-integrality for all $p$, then the (global) integrality follows. $\endgroup$ – WhatsUp Mar 19 at 3:13
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    $\begingroup$ Reminds me of Somos Sequence's. $\endgroup$ – Vepir Mar 19 at 11:27
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    $\begingroup$ @Vepir Probably not surprising because the author of OEIS sequence A248049 is Michael Somos. $\endgroup$ – user Mar 19 at 11:34
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    $\begingroup$ Similar to math.stackexchange.com/questions/1905063/… (note: the proof is less than 100% verified and rather unsatisfactory in its brute-force component). $\endgroup$ – darij grinberg Mar 21 at 15:08

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