4 added 6 characters in body

Let me slightly change the notation. You have some sequence $\mathbf{a}=a_0,a_1,\cdots$ with each $a_i$ either $0$ or $1$. Define a sequence $u_0,u_1,u_2,\cdots$ by setting $u_m=0$ for $m \lt 0$, $u_0=1$ and for any $n \ge 0$, $u_{n+1}=\sum_{i=1}^{\infty}a_iu_{n-i}.$ As you note in a comment, if $a_1=1$ then $u_i$ is a non-decreasing and eventually increasing sequence (excluding a trivial case). The sequence $u_i$ will eventually be increasing as long as the set of $j$ with $a_j=1$ has a $\gcd$ of $1$.

Consider the power series $f(r)=r-(a_0+\frac{a_1}{r}+\frac{a_2}{r^2}+\cdots).$ Then $f(r)$ is defined and increasing for $r \gt 1.$ There is a unique $\lambda=\lambda_{\mathbf{a}}>1$ with $f(\lambda)=0.$ I think that there should be a constant $c \le 1$ with $\lim \frac{a_n}{c\lambda^n}=1$. We can also say that $\lambda \le 2$ with equality only in the degenerate case that $a_i=1$ for all $i$. For any $1 \lt r \lt 2$ we can create an $\mathbf{a}$ with $\lambda_{\mathbf{a}}=r$ by simply choosing the $a_i$ one at a time to keep the (non-decreasing) partial sums of $f(r)$ non-negative.

We can partially order the possible sequences $\mathbf{a}=(a_i)$ by saying $\mathbf{a}<mathbf{b}$<\mathbf{b}$ when$a_i=b_i$for$0 \le i \lt j$and $0=a_j<b_j=1.$ In this case, $\lambda{\mathbf{a}}<\lambda_{\mathbf{b}}$. In case$a_i=0$from some point on we know how to find$\lambda$as the root of a polynomial and if$a_i=1$from some point on we again know how to find$\lambda$as the root of a polynomial (perhaps with a coefficient equal to$2$.So this gives us both a lower bound and an upper bound on$\lambda_{\mathbf{a}}$based on any initial portion of$\mathbf{a}.$I'll stop there for now although probably more can be said. Consider the$2^d$sequences$\mathbf{a}$with$a_{d+m}=a_d$a_{d+m}=0$ for $m \gt 0.$ If all the roots, real and complex , of $f(r)$ are plotted then you should get a picture something like the one below. This plot is for $d=11$ d=12$you can see a similar one at this question . 3 uploaded a picture Let me slightly change the notation. You have some sequence$\mathbf{a}=a_0,a_1,\cdots$with each$a_i$either$0$or$1$. Define a sequence$u_0,u_1,u_2,\cdots$by setting$u_m=0$for$m \lt 0$,$u_0=1$and for any$n \ge 0$,$u_{n+1}=\sum_{i=1}^{\infty}a_iu_{n-i}.$As you note in a comment, if$a_1=1$then$u_i$is a non-decreasing and eventually increasing sequence (excluding a trivial case). The sequence$u_i$will eventually be increasing as long as the set of$j$with$a_j=1$has a$\gcd$of$1$. Consider the power series$f(r)=r-(a_0+\frac{a_1}{r}+\frac{a_2}{r^2}+\cdots).$Then$f(r)$is defined and increasing for$r \gt 1.$There is a unique$\lambda=\lambda_{\mathbf{a}}>1$with$f(\lambda)=0.$I think that there should be a constant$c \le 1$with$\lim \frac{a_n}{c\lambda^n}=1$. We can also say that$\lambda \le 2$with equality only in the degenerate case that$a_i=1$for all$i$. For any$1 \lt r \lt 2$we can create an$\mathbf{a}$with$\lambda_{\mathbf{a}}=r$by simply choosing the$a_i$one at a time to keep the (non-decreasing) partial sums of$f(r)$non-negative. We can partially order the possible sequences$\mathbf{a}=(a_i)$by saying $\mathbf{a}<mathbf{b}$ when$a_i=b_i$for$0 \le i \lt j$and$0=a_j

In case $a_i=0$ from some point on we know how to find $\lambda$ as the root of a polynomial and if $a_i=1$ from some point on we again know how to find $\lambda$ as the root of a polynomial (perhaps with a coefficient equal to $2$.So this gives us both a lower bound and an upper bound on $\lambda_{\mathbf{a}}$ based on any initial portion of $\mathbf{a}.$

I'll stop there for now although probably more can be said.

Consider the $2^d$ sequences $\mathbf{a}$ with $a_{d+m}=a_d$ for $m \gt 0.$ If all the roots, real and complex , of $f(r)$ are plotted then you should get a picture something like the one below. This one comes form polynomials with all coefficients $\pm 1$ . The illustration plot is lifted from this question . For the problem here I would expect more of a C shape around the unit circle with distance from th origin between $\sqrt{2}$ and for $1/\sqrt{2}$. There should also be d=11$you can see a portion of the real line between 1 and 2 along with another portion centered near$-1$on the negative real axissimilar one at this question . 2 rewrite I like Let me slightly change the question, but it needs to be modifiednotation. In the positive direction we can bound You have some sequence$\lambda$in whatever correct result is true as \mathbf{a}=a_0,a_1,\cdots$ with each $u_n a_i$ either $0$ or $1$. Define a sequence $u_0,u_1,u_2,\cdots$ by setting $u_m=0$ for $m \lt 2^n$ 0$,$u_0=1$and for any$n \gt 1.$In the negative directionge 0$, it is easy to arrange for $u_1=u_3=u_5=\cdots=0$ u_{n+1}=\sum_{i=1}^{\infty}a_iu_{n-i}.$As you note in a comment, just set if$a_{2k+1}=0$for all a_1=1$ then $k$ u_i$is a non-decreasing and let the even values eventually increasing sequence (excluding a trivial case). The sequence$a_{2k}$u_i$ will eventually be a mix increasing as long as the set of $0$ and j$with$a_j=1$has a$\gcd$of$1$.To avoid that it might be enough to say that there Consider the power series$f(r)=r-(a_0+\frac{a_1}{r}+\frac{a_2}{r^2}+\cdots).$Then$f(r)$is no arithmetic progression which defined and increasing for$r \gt 1.$There is all a unique$0.$Or one might be able to say \lambda=\lambda_{\mathbf{a}}>1$ with $f(\lambda)=0.$ I think that there are should be a finite number of arithmetic progressions each constant $c \le 1$ with its own growth rate.

Can you give a particular example $\lim \frac{a_n}{c\lambda^n}=1$. We can also say that $\lambda \le 2$ with equality only in the kind of bound degenerate case that you would like? $a_i=1$ for all $i$. For a finite recurrence any $1 \lt r \lt 2$ we can create an $\mathbf{a}$ with $\lambda_{\mathbf{a}}=r$ by simply choosing the $a_i$ one has at a growth rate time to keep the (non-decreasing) partial sums of $f(n)\lambda^n$ f(r)$non-negative. We can partially order the possible sequences$\mathbf{a}=(a_i)$by saying $\mathbf{a}<mathbf{b}$ when$a_i=b_i$for some polynomial$f$0 \le i \lt j$ and constant $\lambda.$ I'm not sure, but I expect that one could choose the 0=a_j

In case $a_i$ a_i=0$from some point on an ad hoc basis we know how to get find$\lambda$as the root of a growth rate polynomial and if$g(n) \lambda^n$a_i=1$ from some point on we again know how to find $\lambda$ as the root of a polynomial (perhaps with a coefficient equal to $g(n)$ 2$.So this gives us both a lower bound and an upper bound on$\lambda_{\mathbf{a}}$based on any initial portion of$\mathbf{a}.$I'll stop there for now although probably more exotic increasing function such as can be said. Consider the$n 2^d$sequences$\mathbf{a}$with$a_{d+m}=a_d$for$m \sqrt{n}.$gt 0.$ If all the roots, real and complex , of $f(r)$ are plotted then you should get a picture something like the one below. This one comes form polynomials with all coefficients $\pm 1$ . The illustration is lifted from this question . For the problem here I would expect more of a C shape around the unit circle with distance from th origin between $\sqrt{2}$ and $1/\sqrt{2}$. There should also be a portion of the real line between 1 and 2 along with another portion centered near $-1$ on the negative real axis.

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