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From This short paper

For polynomial $F$ with integer coefficients, define the recurrence $f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that $f(n)$ satisfy Somos like sequence
$f(n)=\frac{G(f(n-1),...,f(n-d_1))}{H(f(n-1),...,f(n-d_2))}$ for polynomials $G$,$H$ which do not depend on $n$.

The strong conjecture is for all $F$ and the weak conjecture is about $F$ which is linear in $f(n-i)$.

It solves special parametric cases like $f(n)=F(n,f(n-1),f(n-2))=(f(n-1)^2)(b_1 n+b_2)+f(n-2)(b_3 n+b_4)$ for all integers $b_i$ and $f(n)=f(n-1)(n^2+n+2)+f(n-2)(3 n+ 5 )$. The proofs are based on finding algebraic dependency of polynomials.

Q1 Are the conjectures true?

Added later

In comments was suggested existence of algebraic dependency not depending on $n$ using resultants of consecutive terms.

This doesn't prove the conjectures, since additional linearity "luck" is needed.

From Our short paper

For polynomial $F$ with integer coefficients, define the recurrence $f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that $f(n)$ satisfy Somos like sequence
$f(n)=\frac{G(f(n-1),...,f(n-d_1))}{H(f(n-1),...,f(n-d_2))}$ for polynomials $G$,$H$ which do not depend on $n$.

The strong conjecture is for all $F$ and the weak conjecture is about $F$ which is linear in $f(n-i)$.

It solves special parametric cases like $f(n)=F(n,f(n-1),f(n-2))=(f(n-1)^2)(b_1 n+b_2)+f(n-2)(b_3 n+b_4)$ for all integers $b_i$ and $f(n)=f(n-1)(n^2+n+2)+f(n-2)(3 n+ 5 )$. The proofs are based on finding algebraic dependency of polynomials.

Q1 Are the conjectures true?

Added later

In comments was suggested existence of algebraic dependency not depending on $n$ using resultants of consecutive terms.

This doesn't prove the conjectures, since additional linearity "luck" is needed.

From This short paper

For polynomial $F$ with integer coefficients, define the recurrence $f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that $f(n)$ satisfy Somos like sequence
$f(n)=\frac{G(f(n-1),...,f(n-d_1))}{H(f(n-1),...,f(n-d_2))}$ for polynomials $G$,$H$ which do not depend on $n$.

The strong conjecture is for all $F$ and the weak conjecture is about $F$ which is linear in $f(n-i)$.

It solves special parametric cases like $f(n)=F(n,f(n-1),f(n-2))=(f(n-1)^2)(b_1 n+b_2)+f(n-2)(b_3 n+b_4)$ for all integers $b_i$ and $f(n)=f(n-1)(n^2+n+2)+f(n-2)(3 n+ 5 )$. The proofs are based on finding algebraic dependency of polynomials.

Q1 Are the conjectures true?

Added later

In comments Max Alekseyev proves existence of algebraic dependency not depending on $n$ using resultants of consecutive terms.

This is not enough for a proof, because in addition we need the dependency to be linear in the leading term.

The paper discusses this issue and in this case of failure it recommends to increase the "order" of the algebraic dependency.

In a deleted answer Max suggests the counterexample $f(n) = (n+1)^2 f(n-1) + n^2 f(n-2)$, but our sage implementation found closed form of order $5$ using algebraic dependency, disproving the counterexample.

From This short paper

For polynomial $F$ with integer coefficients, define the recurrence $f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that $f(n)$ satisfy Somos like sequence
$f(n)=\frac{G(f(n-1),...,f(n-d_1))}{H(f(n-1),...,f(n-d_2))}$ for polynomials $G$,$H$ which do not depend on $n$.

The strong conjecture is for all $F$ and the weak conjecture is about $F$ which is linear in $f(n-i)$.

It solves special parametric cases like $f(n)=F(n,f(n-1),f(n-2))=(f(n-1)^2)(b_1 n+b_2)+f(n-2)(b_3 n+b_4)$ for all integers $b_i$ and $f(n)=f(n-1)(n^2+n+2)+f(n-2)(3 n+ 5 )$. The proofs are based on finding algebraic dependency of polynomials.

Q1 Are the conjectures true?

Added later

In comments was suggested existence of algebraic dependency not depending on $n$ using resultants of consecutive terms.

This doesn't prove the conjectures, since additional linearity "luck" is needed.

From This short paper

For polynomial $F$ with integer coefficients, define the recurrence $f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that $f(n)$ satisfy Somos like sequence
$f(n)=\frac{G(f(n-1),...,f(n-d_1))}{H(f(n-1),...,f(n-d_2))}$ for polynomials $G$,$H$ which do not depend on $n$.

The strong conjecture is for all $F$ and the weak conjecture is about $F$ which is linear in $f(n-i)$.

It solves special parametric cases like $f(n)=F(n,f(n-1),f(n-2))=(f(n-1)^2)(b_1 n+b_2)+f(n-2)(b_3 n+b_4)$ for all integers $b_i$ and $f(n)=f(n-1)(n^2+n+2)+f(n-2)(3 n+ 5 )$. The proofs are based on finding algebraic dependency of polynomials.

Q1 Are the conjectures true?

From This short paper

For polynomial $F$ with integer coefficients, define the recurrence $f(n)=F(n,f(n-1),f(n-2),...,f(n-d))$. We conjecture that $f(n)$ satisfy Somos like sequence
$f(n)=\frac{G(f(n-1),...,f(n-d_1))}{H(f(n-1),...,f(n-d_2))}$ for polynomials $G$,$H$ which do not depend on $n$.

The strong conjecture is for all $F$ and the weak conjecture is about $F$ which is linear in $f(n-i)$.

It solves special parametric cases like $f(n)=F(n,f(n-1),f(n-2))=(f(n-1)^2)(b_1 n+b_2)+f(n-2)(b_3 n+b_4)$ for all integers $b_i$ and $f(n)=f(n-1)(n^2+n+2)+f(n-2)(3 n+ 5 )$. The proofs are based on finding algebraic dependency of polynomials.

Q1 Are the conjectures true?

Added later

In comments Max Alekseyev proves existence of algebraic dependency not depending on $n$ using resultants of consecutive terms.

This is not enough for a proof, because in addition we need the dependency to be linear in the leading term.

The paper discusses this issue and in this case of failure it recommends to increase the "order" of the algebraic dependency.

In a deleted answer Max suggests the counterexample $f(n) = (n+1)^2 f(n-1) + n^2 f(n-2)$, but our sage implementation found closed form of order $5$ using algebraic dependency, disproving the counterexample.