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\begin{document}
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\begin{center}
\vskip 1cm{\LARGE\bf On Determining Paint by Numbers Puzzles \\
\vskip .08in
with Nonunique Solutions}
\vskip 1cm
\large
Ryan Mullen \\
Sacred Heart University\\
Fairfield, CT 06825 \\
USA \\
\href{mailto:mullenr@sacredheart.edu}{\tt mullenr@sacredheart.edu} \\
\end{center}
\vskip .2 in
\begin{abstract}
{\it Paint by Numbers} is a classic logic puzzle in which the squares of a $p
\times n$ grid are to be colored in such a way as to display a picture.
The decision on which squares to color is determined by sequences of
numbers above each column and to the left of each row. The numbers
describe how many consecutive squares are to be colored in that row or
column, and multiple numbers represent multiple blocks of colored in
squares (with at least one uncolored square inbetween blocks). Certain
natural questions arise. For a given $p \times n$ grid, how many
possible sequences are in a single column or row? For a given grid,
how many puzzles are there? How many of these have unique solutions?
We will explore these questions as well as connections between Paint by
Numbers puzzles, partition theory, and the Fibonacci sequence.
\end{abstract}
\section {Introduction}
{\it Paint by Numbers} (also called Nonograms, Picross, and many other
names) is a logic puzzle which allows purely left brained thinkers to
draw pictures of Matt Damon, Jim Morrison, a shootout at a poker table,
or any picture that can be converted into a two toned highly pixelated
version. The reward for completing the puzzle correctly is a completed
drawing; however, solving the puzzle needs no greater artistic ability
than staying inside the lines. Ueda and Nagao \cite{np} have shown that
determining if a solution is unique is NP-complete (a famous complexity class
from computer science). Batenburg and Kosters
\cite{solving} have shown that under certain assumptions puzzles can be
solved in polynomial time. In addition Benton, Snow, and
Wallach \cite{rowsums} compute the number of possible row and column
sums.
The puzzle starts out with a $p\times n$ grid with sequences of numbers above each column and sequences of numbers alongside each row. The numbers tell you how many consecutive squares in the column or row need to be colored in. If there is more than one number in the sequence then there must be at least one uncolored (white) space between the colored (black) spaces. Figure \ref{start} is an example of a small puzzle.
\begin{figure}[h!t!]
\begin{center}
\begin{tabular}{c|c|c|c|c|c|}
&&2&&2&\\
&1&1&5&1&1\\
\hline
1&&&&&\\
\hline
3&&&&&\\
\hline
3&&&&&\\
\hline
1&&&&&\\
\hline
5&&&&&\\
\hline
\end{tabular}
\end{center}
\caption{Example of a small puzzle}
\label{start}
\end{figure}
Initially we see that along the bottom and in the middle column we must have 5 black squares, and there is exactly room for 5. Thus we can fill in the following:
\begin{figure}[h!t!]
\begin{center}
\begin{tabular}{c|c|c|c|c|c|}
&&2&&2&\\
&1&1&5&1&1\\
\hline
1&&&\sbl&&\\
\hline
3&&&\sbl&&\\
\hline
3&&&\sbl&&\\
\hline
1&&&\sbl&&\\
\hline
5&\sbl&\sbl&\sbl&\sbl&\sbl\\
\hline
\end{tabular}
\end{center}
\caption{Intermediate stage}
\label{middle}
\end{figure}
Next we note that the top row is completed. That leaves only four remaining squares in which to fill columns 2 and 4 and that is just barely enough. See Figure \ref{finish}.
\begin{figure}[h!t!]
\begin{center}
\begin{tabular}{c|c|c|c|c|c|}
&&2&&2&\\
&1&1&5&1&1\\
\hline
1&&&\sbl&&\\
\hline
3&&\sbl&\sbl&\sbl&\\
\hline
3&&\sbl&\sbl&\sbl&\\
\hline
1&&&\sbl&&\\
\hline
5&\sbl&\sbl&\sbl&\sbl&\sbl\\
\hline
\end{tabular}
\end{center}
\caption{Finished picture}
\label{finish}
\end{figure}
The puzzle is complete, and we have ``drawn" a lonely tree.
If a sequence such as $3,1,4$ is listed alongside a row of the puzzle, then we know that there must be 3 black squares, at least one white, one black square, at least one white, and then 4 black squares. There may or may not be white squares before the 3 black and after the 4 black. The order is important as $4,1,3$ would describe a different coloring. Depending on how large the puzzle is, we may or may not be sure certain squares are colored. For example if $n=10$ then we know exactly which squares are to be colored (see Figure \ref{force10}).
\begin{figure}[h!t!]
\begin{center}
\bl\bl\bl\wh\bl\wh\bl\bl\bl\bl
\end{center}
\caption{Forced coloring with 10 squares}
\label{force10}
\end{figure}
If $n=11$, then there are four possible configurations; however all four have the black squares in Figure \ref{force11}.
\begin{figure}[h!t!]
\begin{center}
\wh\bl\bl\wh\wh\wh\wh\bl\bl\bl\wh
\end{center}
\caption{Forced coloring with 11 squares}
\label{force11}
\end{figure}
Filling in the remaining black squares can be done using information from the columns and the other rows. If $n=14$ or more, then we would not be able to fill in any of the squares in this row without using information from the rest of the puzzle because this sequence uses 10 squares and there would be at least 4 left over. That leaves the largest piece of 4 with two nonoverlapping possible locations. We define a {\it forceless} sequence to be a sequence such that no partial coloring can be determined without the aid of the rest of the puzzle (see Figure \ref{extreme}).
\section{Number of possible sequences}
For any $p\times n$ puzzle there are $p$ sequences of natural numbers one to the left of each row and there are $n$ sequences of natural numbers one above each column. For a row sequence $\left\{a_j\right\}_{j=1}^i=\mathbf{a_j}$ we define $l(\mathbf{a_j})$ to be the {\it length} of the sequence, which is the fewest number of columns needed for that sequence to fit in a puzzle. Since there must be at least one white square between each string of black squares the formula is
$$l(\mathbf{a_j})=\sum_{j=1}^ia_j+i-1.$$
Also define the norm of a row sequence as
$$\|\mathbf{a_j}\|=\max_{1\le j\le i}a_j.$$
A row sequence $\mathbf{a_j}$ in a puzzle with $n$ columns would be forceless if $\mathbf{a_j}\ne\{\}$ and
$$l(\mathbf{a_j})+\|\mathbf{a_j}\|\le n.$$
Column sequences have corresponding definitions but we will focus on row sequences. For a given $n$, just how many sequences are allowable? Somewhat surprisingly the number of sequences follows a well known sequence. The Fibonacci sequence $F_n$ (A000045) is determined recursively as
$$F_1=F_2=1;$$
$$F_n=F_{n-1}+F_{n-2}.$$
\begin{theorem}
For a puzzle with $n$ columns, the number of row sequences that describe a coloring which fits in the puzzle is $F_{n+2}.$
\end{theorem}
\begin{proof}
Let $b_n$ be the number of sequences in a puzzle with $n$ columns. For the case $n=1$, there are 2 possible sequences: $\{\}$ and $1$. Thus $b_1=2$. For the case $n=2$, there are 3 possible sequences: $\{\}$, $1$, and $2$. Thus $b_2=3$. If $n>2$, the number of sequences allowed is $b_{n-1}$ plus the number of sequences of length $n$. If a sequence has length $n$ and has $i$ entries, then there must be $i-1$ white squares and $n-i+1$ black squares. To count the number of sequences, think of starting with a long string of $n-i+1$ black squares and then choosing $i-1$ places to cut the string of black squares and placing a white squares in the cuts. We can only cut between black squares and not in the middle of any square. Thus there are $n-i$ possible cutting points and $i-1$ cuts needed to be made, and the cutting order is irrelevent which gives
$\left(n-i\atop{i-1}\right)$ sequences of $i$ numbers of length $n$. Thus
\begin{align*}
b_n&=b_{n-1}+\sum_{i=1}^{\lfloor\f{n+1}{2}\rfloor}\left(n-i\atop{i-1}\right)\\
&=b_{n-1}+\sum_{i=0}^{\lfloor\f{n-1}{2}\rfloor}\left(n-i-1\atop{i}\right)\\
&=b_{n-1}+1+\sum_{i=1}^{\lfloor\f{n-1}{2}\rfloor}\left(n-i-2\atop{i}\right) +\left(n-i-2\atop{i-1}\right)\\
&=b_{n-1}+\sum_{i=1}^{\lfloor\f{n}{2}\rfloor}\left(n-i-1\atop{i-1}\right)+\sum_{i=1}^{\lfloor\f{n-1}{2}\rfloor}\left(n-i-2\atop{i-1}\right)\\
&=2b_{n-1}-b_{n-3}.
\end{align*}
If we assume $b_{n-1}=b_{n-2}+b_{n-3}$ and use induction, we have
\begin{align*}
b_n&=b_{n-1}+b_{n-2}+b_{n-3}-b_{n-3}\\
&=b_{n-1}+b_{n-2}.
\end{align*}
So the sequence $b_n$ has the same recursion as the Fibonacci sequence but is shifted since $b_n$ starts with 2 and 3 rather than 1 and 1. So $b_n=F_{n+2}$.
\end{proof}
This proof also illustrates the closed form formula for the Fibonacci numbers
$$F_n=\sum_{i=1}^{\lfloor\f{n+1}{2}\rfloor}\left(n-i\atop{i-1}\right).$$
When counting the number of sequences of length $n$, we are almost counting the partitions of $n-i$ into $i-1$ pieces and to get all the partitions of $n-i$ we could simply sum over all possible numbers of pieces. If it were not for the seemingly trivial detail that some of the partitions could be counted numerous times, this would work. For example the partition $\{1,1,2\}$ of $5$ would get counted 3 times since $(2,1,1), (1,2,1)$ and $(1,1,2)$ are different row sequences.
\section{Forceless sequences}
In the process of solving a Paint by Numbers puzzle, the first thing to look for are sequences that force a coloring. But just how rare is a sequence with no forcing? A sequence would be forceless if in at least two possible colorings, the largest block of squares are nonoverlapping. In Figure \ref{extreme} we see that the largest block of black squares, 4, has no overlapping between the extreme left and extreme right color placements. Thus $3,1,4$ would be a forceless row sequence in a puzzle with 14 columns. The goal of the remainder of the paper is to count the number of forceless sequences.
\begin{figure}[h!t!]
\begin{center}
3 1 4 \bl\bl\bl\wh\bl\wh\bl\bl\bl\bl\wh\wh\wh\wh
\end{center}
\vspace{10pt}
\begin{center}
3 1 4 \wh\wh\wh\wh\bl\bl\bl\wh\bl\wh\bl\bl\bl\bl
\end{center}
\caption{Extreme left and extreme right of color placement}
\label{extreme}
\end{figure}
In general, we get a forceless sequence in a puzzle with $n$ columns if the row sequence $\{a_j\}$ satisfies
$$\|{\mathbf a_j}\|+l({\mathbf a_j})\le n.$$
For the moment, let's consider sequences with a fixed length $m$. If this sequence is to be forceless, then its maximum norm is $n-m$. Let $A_{m,k}$ denote the sequences of length $m$ and maximum norm $k$. To count all forceless sequences in a puzzle with $n$ columns we need only sum $A_{m,n-m}$ over all possible lengths. That is, the number of forceless sequences, denoted $G_n$, is
$$G_n=\sum_{m=0}^{n-1} A_{m,n-m}.$$
To calculate $A_{m,k}$, it is easier to fix the maximum norm and let the length vary. We start with $k=1$. Since the maximum norm is one, the sequence must consist entirely of ones. Say there are $r$ ones. The length of a sequence consisting entirely of ones is $l({\mathbf a_j})=r+r-1=2r-1$. Therefore if $m$ is odd, there is one sequence with maximum norm 1, and if $m$ is even, there are no such sequences. That is $A_{m,1}=1$ for $m$ odd, and $A_{m,1}=0$ for $m$ even.
Moving on to $k=2$, there is only the sequence $\{1\}$ of length $1$ and there is only the sequence $\{2\}$ of length $2$. So $A_{1,2}=A_{2,2}=1$. Considering $A_{m,2}$, we are counting the number of sequences of length $m$ and maximum norm 2. There are two ways these sequences could come about. Either we addend a $1$ to a sequence of length $m-2$ or addend a $2$ to a sequence of length $m-3$. Therefore $A_{m,2}=A_{m-2,2}+A_{m-3,2}$.
For a general $k$, similar reasoning shows $A_{1,k}=A_{2,k}=1$. For $m\le k$, the sequence $\{m\}$ has length $m$ and max norm less than or equal to $k$, and is not obtained from addending a number to a previously counted sequence. However, all other sequences are obtained in this manner. The remaining sequences of length $m$ are obtained by addending a 1 to the sequences of length $m-2$, addending a 2 to the sequences of length $m-3$,..., or addending a $m-2$ to the sequence of length 1. If $m>k$, then the sequence $\{m\}$ has a norm which is too large; thus that sequence would not be counted. So all sequences are obtained by addending a number to a previous sequence. The sequences of length $m$ are obtained by addending a 1 to the sequences of length $m-2$, addending a 2 to the sequences of length $m-3$,..., or addending an $m$ to the sequences of length $m-k-1$. From this we obtain the following formula:
$$A_{m,k}=\left\{\begin{array}{cc} 1+\sum_{j=1}^{m-2} A_{j,k},&m\le k;\\ \sum_{j=m-k-1}^{m-2}A_{j,k},&m>k.\end{array}\right.$$
Simple induction arguments can be used to obtain
$$A_{m,k}=\left\{\begin{array}{cc} F_m,&m\le k;\\ A_{m-1,k}+A_{m-2,k}-A_{m-k-2,k}, &m>k.\end{array}\right.$$
For $k=1,\ldots,9$, these sequences are in the On-Line Encyclopedia of Integer Sequences, $A_{m,1}$ (A000035), $A_{m,2}$ (A000931), $A_{m,3}$ (A013979), $A_{m,4},\ldots,A_{m,9}$ (A013982),\ldots,(A013987).
\begin{table}
\begin{center}
\begin{tabular}{c|cccccccccc}
$m,k\to$&1&2&3&4&5&6&7&8&9&10\\
\hline
1&1&1&1&1&1&1&1&1&1&1\\
2&0&1&1&1&1&1&1&1&1&1\\
3&1&1&2&2&2&2&2&2&2&2\\
4&0&2&2&3&3&3&3&3&3&3\\
5&1&2&4&4&5&5&5&5&5&5\\
6&0&3&5&7&7&8&8&8&8&8\\
7&1&4&8&10&12&12&13&13&13&13\\
8&0&5&11&16&18&20&20&21&21&21\\
9&1&7&17&24&29&31&33&33&34&34\\
10&0&9&24&37&45&50&52&54&54&55\\
\end{tabular}
\end{center}
\caption{$A_{m,k}$ for $1\le m,k \le 10$}
\label{amk}
\end{table}
It is worth noting that for every fixed $m$, $A_{m,k}$ is an increasing sequence in $k$. In other words, if the length is fixed and the maximum norm is allowed to increase, more sequences would be obtained. Also for every fixed $k>1$, $A_{m,k}$ is an increasing sequence in $m$. Also notice that if we fix a length $m$ and consider the sequence in that row, the sequence is bounded. In fact, it is bounded by the $m$th Fibonacci number. This should by no means be surprising since $m$ represents the length of sequence, and as we increase the maximum norm (largest piece size) we will eventually get all sequences of length $m$, which we have already decided is determined by the Fibonacci sequence.
To obtain $G_n$, we simply add all terms such that $m+k=n$; that is, add along diagonals in Table \ref{amk}. Table \ref{firstterms} gives the values of $G_n$ up to $n=11$.
\begin{table}[h!t!]
\begin{align*}
G_2&=1\\
G_3&=0+1=1\\
G_4&=1+1+1=3\\
G_5&=0+1+1+1=3\\
G_6&=1+2+2+1+1=7\\
G_7&=8\\
G_8&=15\\
G_9&=20\\
G_{10}&=33\\
G_{11}&=47
\end{align*}
\caption{$G_n$ for small $n$}
\label{firstterms}
\end{table}
So far we have outlined a way of computing the number of forceless row sequences for any number of columns $n$. It would be nice to have a closed form formula to do this more efficiently. That goal seems unlikely to be met, however other questions can be answered about $\{G_n\}.$ In the remainder of the paper we will discuss the growth rate of $\{G_n\}$ and how that growth relates to the growth rate of the Fibonacci sequence. In particular we will prove the following results.
\begin{theorem} \label{lowerbound} For every $n$, $G_n\ge C_1 \f{\phi^n}{n}$, where $\phi$ is the golden ratio.
\end{theorem}
\begin{theorem} \label{upperbound} For every $n$, $G_n\le C \f{\ln n \phi^n}{n}.$
\end{theorem}
But first we present an alternate method for computing the sequence $\{G_n\}$.
\section{Generating polynomial}
We can also use polynomials to find the sequence $\{G_n\}$. Consider the polynomial \begin{equation}\label{poly}x^i\left(\sum_{j=1}^k x^j\right)^{i+1}.\end{equation} For a fixed $k$, let us examine this polynomial as $i$ goes from 0 to $\infty$. When $i=0$, the coefficient of $x^m$ in (\ref{poly}) is the number of ways to pick one number between 1 and $k$ with sum $m$. In the context of Paint by Numbers puzzles, the coefficient of $x^m$ is the number of sequences with norm $k$, length $m$, and one entry. When $i=1$, the coefficient of $x^{m-1}$ in $\left(\sum_{j=1}^k x^j\right)^{i+1}$ is the number of ways to pick two numbers between 1 and $k$ with sum $m-1$. Again in our context, we have the number of sequences with norm $k$, length $m$, and two entries. Thus the coefficient of $x^m$ in (\ref{poly}) is the number of sequences with norm $k$, length $m$, and two entries. For a general $i$, the coefficient of $x^{m-i}$ in $\left(\sum_{j=1}^k x^j\right)^{i+1}$ is the number of sequences with norm $k$, length $m$, and $i+1$ entries. Hence the coefficient of $x^m$ in (\ref{poly}) is the number of sequences with norm $k$, length $m$ and $j+1$ pieces. If we sum over all possible numbers of pieces, we will have all sequences of length $m$ and norm $k$. Therefore,
\begin{equation}\label{poly2} \sum_{m=1}^\infty x^mA_{m,k}=\sum_{j=0}^\infty x^j\left(\sum_{i=1}^k x^i\right)^{j+1}.\end{equation}
Now we compute the generating polynomial for $G_n$ as follows:
\begin{align*}
\sum_{n=1}^\infty G_n x^n&=\sum_{n=1}^\infty x^n \sum_{k=1}^{n-1}A_{n-k,k}\\
&=\sum_{k=1}^\infty \sum_{n=k+1}^\infty x^n A_{n-k,k}\\
&=\sum_{k=1}^\infty x^k\sum_{m=1}^\infty x^m A_{m,k}\\
&=\sum_{k=1}^\infty x^k\sum_{j=0}^\infty x^j \left(\sum_{i=1}^kx^i\right)^{j+1}.
\end{align*}
\section{Lower bound for $G_n$}
\subsection{The main idea}
Unfortunately $G_n$ is not itself a linear recursion so finding a closed form formula for it is a daunting task. However, the columns (fix $k$ and consider the sequence $A_{m,k}$ in $m$) are linear recursions. We will use this fact along with some clever observations to find asymptotic results for $G_n$.
First let us find a lower bound for $G_n$. When calculating $G_{11}$, we add the red numbers in Table \ref{lowerg11}.
\begin{table}
\begin{center}
\begin{tabular}{c|cccccccccc}
$m,k\to$&1&2&3&4&5&6&7&8&9&10\\
\hline
1&1&1&1&1&1&1&1&1&1&{\color{red}1}\\
2&0&1&1&1&1&1&1&1&{\color{red}1}&1\\
3&1&1&2&2&2&2&2&{\color{red}2}&2&2\\
4&0&2&2&3&3&3&{\color{red}3}&3&3&3\\
5&1&2&4&4&5&{\color{red}5}&5&5&5&5\\
6&0&3&5&7&{\color{red}7}&8&8&8&8&8\\
7&1&4&8&{\color{red}10}&12&12&13&13&13&13\\
8&0&5&{\color{red}11}&16&18&20&20&21&21&21\\
9&1&{\color{red}7}&17&24&29&31&33&33&34&34\\
10&{\color{red}0}&9&24&37&45&50&52&54&54&55\\
\end{tabular}
\end{center}
\caption{$A_{m,k}$ for $1\le m,k \le 10$, red numbers sum to $G_{11}$}
\label{lowerg11}
\end{table}
Now choose any column and add the red number and all numbers above it in that column. No matter which column you choose you will obtain a smaller result than $G_{11}$. This is the idea behind the following proposition.
\begin{proposition}\label{anyk}
For every $k$, $G_m\ge\sum_{j=1}^{m-k}A_{j,k}.$
\end{proposition}
\begin{proof}
Fix $k\ge1$. If $m-j\ge k$, then $A_{j,m-j}\ge A_{j,k}$. This is because for a fixed length $j$, as the maximum norm increases, so does the number of sequences. Therefore
\begin{align*}G_m&=\sum_{j=1}^{m-k}A_{j,m-j}+\sum_{j=m-k+1}^{m-1}A_{j,m-j}\\
&\ge \sum_{j=1}^{m-k}A_{j,k}+0.\\
\end{align*}
\end{proof}
\subsection{Linear recursions}
When studying linear recursions such as the Fibonacci sequence, we notice that the ratios of consecutive terms seem to approach a fixed number (call it the {\it limiting ratio}). If the ratios of consecutive terms were a fixed number $\rho$ then then $m$th term of the sequence would be $\rho^m$ and the sequence would also have to satisfy the Fibonacci recursion
$$\rho^m=\rho^{m-1}+\rho^{m-2}.$$
If we factor out $\rho^{m-2}$, we the get characteristic equation for the Fibonacci recursion $$\rho^2=\rho+1.$$
Solving for $\rho$, we obtain two roots $\phi=\f{1+\sqrt{5}}{2}$ and $\lambda_1=\f{1-\sqrt{5}}{2}$. We find another closed form formula for the Fibonacci sequence by writing $$F_n=b_0\phi^n+b_1\lambda_1^n,$$ and then using the fact that $F_1=1,F_2=1$ to solve for the constants $b_0$ and $b_1$, we obtain $$F_n=\f{\phi^n-\lambda_1^n}{\sqrt{5}}.$$ Notice since $\phi$ is the larger of the two roots (in absolute value), as $n$ increases the Fibonacci numbers grow very closely to $\f{\phi^n}{\sqrt{5}}.$
Just as with all linear recursions, for a fixed $k$, the terms of $A_{m,k}$ can be expressed in a closed form manner using the zeros of the recursion's characteristic polynomial. In this case the characteristic polynomial is
$$x^{k+2}-x^{k+1}-x^k+1.$$ This polynomial has $k+2$ zeros (counting multiplicity), and for $k>1$, one of these zeros is real and larger than 1. We will call this zero $\phi_k$ with multiplicity $\alpha_0$. Label the remaining zeros as $\lambda_1,\ldots,\lambda_j$ with multiplicities $\alpha_1,\ldots,\alpha_j$ (most of these are complex numbers). Using linear recursion theory, we can write
\begin{equation}\label{closed}A_{m,k}=\sum_{i=1}^{\alpha_0}b_{i,0}m^{i-1}\phi_k^m+\sum_{l=1}^j\sum_{i=1}^{\alpha_j}b_{i,l}m^{i-1}\lambda_l^m.\end{equation} The coefficients $b_{i,l}$ can be determined using the first $k+2$ terms of the recursion. So in theory, nice formulas can be obtained for $A_{m,k}$; however computing the zeros and then solving the systems of linear equations can only be done approximately. Nontheless determining which zero is the largest in absolute value will go a long way in approximating $A_{m,k}.$
\begin{proposition}\label{biggest}
Let $k>1$ and $\phi_k,\lambda_1,..,\lambda_j$ be the zeros of $f_k(z)=z^{k+2}-z^{k+1}-z^k+1$, where $\phi_k$ is the real zero that is larger than 1. For all $1\le i\le j$, $\phi_k>|\lambda_i|$. Furthermore, $\phi_k$ is a zero of multiplicity one.
\end{proposition}
\begin{proof}
Since $\phi_k$ is a zero of $f_k(z)$, $(z-\phi_k)$ is a factor. A simple multiplication shows that $$f_k(z)=(z-\phi_k)(z^{k+1}+(\phi_k-1)z^k-\sum_{j=0}^{k-1}z^j\phi_k^{-j-1}).$$
Consider the function $g_k(z)=(\phi_k-1)z^k-\sum_{j=0}^{k-1}z^j\phi_k^{-j-1}$. For all $z$ such that $|z|=\phi_k$, we have
\begin{align*}
|g_k(z)|&\le(\phi_k-1)\phi_k^k+\sum_{j=0}^{k-1}\phi_k^j\phi_k^{-j-1}\\
&=\phi_k^{k+1}-\phi_k^k+\sum_{j=0}^{k-1}\phi_k^{-1}\\
&=\phi_k^{k+1}-\phi_k^k+k\phi_k^{-1}\\
&< \phi_k^{k+1}.
\end{align*}
Where the last inequality is justified because for all $k\ge2$, $\phi_k\ge\phi_2>\sqrt[3]{2}\ge\sqrt[k+1]{k}$.
On the closed curve $|z|=\phi_k$, we have $|g_k(z)|\le \phi_k^{k+1}=|z^{k+1}|$. Thus Rouche's theorem (see any complex analysis textbook, for example \cite{book}) tells us that $z^{k+1}$ and $z^{k+1}+g_k(z)$ have the same number of zeros inside the curve $|z|=\phi_k$. The polynomial $z^{k+1}$ has $k+1$ zeros inside this curve. Therefore all $k+1$ zeros of $z^{k+1}+g_k(z)$ are inside this curve. That is, for all $1\le i\le j$, $\phi_k>|\lambda_i|$ and $\phi_k$ has multiplicity one.
\end{proof}
In light of (\ref{closed}) and Proposition \ref{biggest}, we see that for large $m$ the term $b_{1,0}\phi_k^m$ is the dominating factor in computing $A_{m,k}$. Furthermore $$\lim_{m\to\infty}\f{A_{m+1,k}}{A_{m,k}}=\phi_k.$$
Clearly, $\phi_k<\phi$ for all $k$; the sequences $A_{m,k}$ grow slower than the Fibonacci sequence. Also $\lim_{k\to\infty} \phi_k =\phi$; as $k$ goes to infinity $A_{m,k}$ goes to the Fibonacci sequence. Furthermore, for $k_11$ as well by defining $\phi_k$ to be the largest positive real zero of $f_k(z)=z^{k+2}-z^{k+1}-z^k+1.$ Now that we know which zero to focus on, we must find a way to approximate it. We begin the following subsection with a lower bound for $\phi_k$.
\subsection{Proof of Theorem~\ref{lowerbound}}
\begin{proposition}\label{lower}
For every $k\ge8$, $\phi_k\ge\phi\left(\f{\sqrt{5}\phi^{k+1}-(k+2)}{\sqrt{5}\phi^{k+1}-(k+1)}\right)$.
\end{proposition}
\begin{proof}
Let $f_k(x)=x^2+x-1-x^{k+2}$. The largest zero of $f_k(x)$ will be the reciprocal of $\phi_k$. We use a linear approximation to $f_k$ at $\phi^{-1}$ and the following:
$$f_k(\phi^{-1})=-\phi^{-k-2}$$
$$f'_k(\phi^{-1})=\sqrt{5}-(k+2)\phi^{-k-1}$$
$$f''_k(x)=2-(k+2)(k+1)x^k.$$
So the linear approximation is
$$f_k(x)\approx (\sqrt{5}-(k+2)\phi^{-k-1})(x-\phi^{-1})-\phi^{-k-2}.$$
For $k\ge8$, $2-(k+1)(k+2)\phi^{-k}>0$. Thus $f_k(x)$ is concave up at $x=\phi^{-1}$ and the linear aproximation is an underestimate. Now because the linear approximation is an underestimate and $f_k(\phi^{-1})$ is negative, the approximation of $\phi_k^{-1}$,call it $\widetilde{\phi}_k^{-1}$, using Newton's method, will be larger than $\phi_k^{-1}$. Indeed,
\begin{align*}
\tilde{\phi}_k^{-1}&=\phi^{-1}+\f{\phi^{-k-2}}{\sqrt{5}-(k+2)\phi^{-k-1}}\\
&=\f{\phi^{-1}(\sqrt{5}-(k+2)\phi^{-k-1})+\phi^{-k-2}}{\sqrt{5}-(k+2)\phi^{-k-1}}\\
&=\phi^{-1}\left(\f{\sqrt{5}-(k+1)\phi^{-k-1}}{\sqrt{5}-(k+2)\phi^{-k-1}}\right)\\
&=\phi^{-1}\left(\f{\phi^{k+1}\sqrt{5}-(k+1)}{\phi^{k+1}\sqrt{5}-(k+2)}\right).\\
\end{align*}
Since $\phi_k^{-1}<\widetilde{\phi}_k^{-1}$, we have
$$\phi_k>\phi\left(\f{\phi^{k+1}\sqrt{5}-(k+2)}{\phi^{k+1}\sqrt{5}-(k+1)}\right).$$
\end{proof}
Since $G_n$ is bounded below by the sum of the first few terms of a linear recursion using Proposition \ref{anyk} and in turn those terms are approximately exponential, we can approximate $G_n$ by summing a geometric series. In order to provide the details, a lemma is in order.
\begin{lemma}\label{rhorisrho}
Let $r=\log_{\phi}m$. For all $m>2$, $\phi_r^m\ge C\phi^m$.
\end{lemma}
\begin{proof}
If $m\ge50$, then $r\ge8$, we can use Proposition \ref{lower} to see that
\begin{align*}
\ln\left(\f{\phi_r}{\phi}\right)^m&\ge m \ln \left(\f{\sqrt{5}m\phi-r-2}{\sqrt{5}m\phi-r-1}\right)\\
&=m\ln\left(1-\f{1}{\sqrt{5}m\phi-r-1}\right)\\
&\ge -2m\left(\f{1}{\sqrt{5}m\phi-r-1}\right)\\
&\ge -2m\left(\f{1}{\sqrt{5}m\phi-m}\right)\\
&= \left(\f{-2}{\sqrt{5}\phi-1}\right).\\
\end{align*}
In the third step, we observed that for all $-\f{1}{2}2x$. Exponentiating, we have
$$\left(\f{\phi_r}{\phi}\right)^m\ge e^{\left(\f{-2}{\sqrt{5}\phi-1}\right)}.$$
If $m<50$ then we can bound $\phi_r^m\ge \left(\f{\phi_1}{\phi}\right)^{50}\phi^m$, and the result is obtained with $C=\min\left\{e^{\left(\f{-2}{\sqrt{5}\phi-1}\right)},\left(\f{\phi_1}{\phi}\right)^{50}\right\}.$
\end{proof}
We are now ready to prove Theorem \ref{lowerbound}.
\begin{proof} Let $r=\lfloor \log_\phi n \rfloor$. Using Proposition \ref{anyk} with $k=r$, we have $G_n\ge \sum_{m=1}^{n-r} A_{m,r}$. Since $A_{m,r}$ is a linear recursion with limiting ratio $\phi_r$, $A_{m,r}\ge K_1\phi_r^m$ for some constant $K_1$ which is independent of $m$. Therefore,
\begin{align*}
G_n\ge& \sum_{m=1}^{n-r} A_{m,r}\\
\ge&\sum_{m=1}^{n-r} K_1\phi_{r}^m\\
=& K_2 \phi_{r}^{n-r}\\
\ge& K_2\phi_{\log_\phi n}^{n-r}\\
\ge& K_2\phi_{\log_\phi n}^{n-\log_\phi n-1}\\
\ge& K_2C\phi^{n-\log_\phi n-1}\\
=&\f{C_1\phi^n}{n},
\end{align*}
where in the 6th step we used Lemma \ref{rhorisrho}.
\end{proof}
\section{Upper bound for $G_n$}
\subsection{The main idea}
Let's refer to Table \ref{upperg11} to try to obtain an upper bound for $G_n$. Again the values in red are summed to get $G_{11}$. The important thing to notice for an upper bound is that roughly the first half (starting in the upper right corner) of the numbers to be summed are the Fibonacci sequence. So if we go to the column which starts with the same number of terms from the Fibonacci sequence and sum down, we will have an upper bound. How far down should we sum? Well $G_{11}$ is the sum of 10 terms so using the first 10 terms in that column (underlined numbers in Table \ref{upperg11}) will definitely give an upper bound.
\begin{table}
\begin{center}
\begin{tabular}{c|cccccccccc}
$m,k\to$&1&2&3&4&5&6&7&8&9&10\\
\hline
1&1&1&1&1&1&{\underline 1}&1&1&1&{\color{red}1}\\
2&0&1&1&1&1&{\underline 1}&1&1&{\color{red}1}&1\\
3&1&1&2&2&2&{\underline 2}&2&{\color{red}2}&2&2\\
4&0&2&2&3&3&{\underline 3}&{\color{red}3}&3&3&3\\
5&1&2&4&4&5&{\color{red} \underline{5}}&5&5&5&5\\
6&0&3&5&7&{\color{red}7}&{\underline 8}&8&8&8&8\\
7&1&4&8&{\color{red}10}&12&\underline {12}&13&13&13&13\\
8&0&5&{\color{red}11}&16&18&\underline {20}&20&21&21&21\\
9&1&{\color{red}7}&17&24&29&\underline {31}&33&33&34&34\\
10&{\color{red}0}&9&24&37&45&\underline {50}&52&54&54&55\\
\end{tabular}
\end{center}
\caption{$A_{m,k}$ for $1\le m,k \le 10$, underlined numbers are an upper bound for $G_{11}$}
\label{upperg11}
\end{table}
However, we can find a better upper bound. Towards the end of the column with underlined numbers, the underlined numbers (approximation of $G_{11}$) are much larger than the numbers in red ($G_{11}$). How can we get around using those extraordinarily large numbers? The largest summand used to determine $G_{11}$ is 11. So instead of adding 20, 31, and 50, replace each of those with 11 (notice also that leaving the 20 alone and replacing 31 and 50 with 7 and 7 would give another upper bound). In fact, we need not be concerned about which column we choose. We are finding an upper bound so that we can approximate the first segment of summands with Fibonacci numbers. So for a general $G_n$, the strategy is to locate the largest summand, $g_{\max}$, or at least find a summand so that all the summands to the left and below are smaller. Starting from the upper right, replace all summands in $G_n$ up to $g_{\max}$ with the Fibonacci sequence and replace the remaining summands with $g_{\max}$. But how do we find $g_{\max}$ for an arbitrary $n$? We need to determine which column contains the largest summand for $G_n$. In order to do this we must first find an upper bound for $\phi_k$.
\subsection{Upper bound for $\phi_k$}
\begin{proposition}\label{upper}
For all $k$,
\begin{equation}\label{root}\phiki\ge \f{1+k(k+2)\phi^{-k-1}-\sqrt{5-k(k+2)\phi^{-2k-2}+2\phi^{-k}(k(\phi-3)-2)}}{-2+(k+1)(k+2)\phi^{-k}}.\end{equation}
\end{proposition}
\begin{proof}
As in Proposition \ref{lower}, we will come up with this estimate using a degree 2 Taylor polynomial approximation, $p_k(x)$ to $f_k(x)$ centered at $\phi^{-1}$. The concavity of the approximation is $p''_k(x)=f''_k(\phi^{-1})$, a constant. The concavity of $f_k(x)$ is given by $f''_k(x)=2-(k+1)(k+2)x^k$. If $x>\phi^{-1}$, then the concavity of $f_k$ is less than the concavity of $p_k$. Therefore the slopes of $p_k$ are increasing at a faster rate. Now $f_k(\phi^{-1})=p_k(\phi^{-1})<0$, and $p_k$ increases faster than $f_k$, so the zero of $p_k$, $\widetilde{\phi}^{-1}_k$, will be less than the zero of $f_k$, $\phiki$.
The second degree Taylor polynomial centered at $\phii$ is given by
\begin{align*}
p_k(x)&=\f{f_k''(\phii)}{2}x^2+x\left(f'_k(\phii)-f''_k(\phii)\phii\right)+\left(f_k(\phii)-f'_k(\phii)\phii+\f{f_k''(\phii)}{2}\phi^{-2}\right)\\
&=ax^2+bx+c,
\end{align*}
where
\begin{align*}
a&=1-\f{(k+1)(k+2)}{2}\phi^{-k},\\
b&=1+k(k+2)\phi^{-k-1},\\
c&=-1-\f{k(k+1)}{2}\phi^{-k-2}.
\end{align*}
The approximation $p_k(x)$ will have two roots the positive one is the right hand side of (\ref{root}). To find the zeros of $p_k$, we use the quadratic formula and much omitted algebra.
\end{proof}
The upper bound for $\phi_k$ (as well as the lower bound) appear to be artificially complicated. They are only bounds, after all. Why not find simpler looking bounds? For example, why not use \begin{equation}\label{badest}\phiki\ge \f{1+k(k+2)\phi^{-k-1}-\sqrt{5}}{-2+(k+1)(k+2)\phi^{-k}}\end{equation} to obtain an upper bound for $\phi_k$? We will see later that every term in the bounds from Propositions \ref{lower} and \ref{upper} is necessary for later estimates to work out.
\subsection{Determing how to split the sum}
It turns out that when finding an upper bound for $G_n$, locating the largest summand, $g_{\max}$ is not crucial. We will estimate $G_n$ by splitting the sum into two pieces. We can split the sum anywhere as long as the summand before (left and below) the split is larger than any other summand before the split. In other words, we need to find a number $k_s$ such that, $A_{n-k_s,k_s}>A_{n-k,k}$ for all $k\phi_k$, if $n\ge\f{\ln\phi_{k+1}}{\ln\f{\phi_{k+1}}{\phi_k}}$, then $\phi_k^n\le\phi_{k+1}^{n-1}$. So we only need find an upper bound for $\f{\ln\phi_{k+1}}{\ln\f{\phi_{k+1}}{\phi_k}}$. Let us start by finding a lower bound for $\ln\f{\phi_{k+1}}{\phi_k}.$ Indeed,
\begin{align*}
\ln\f{\phi_{k+1}}{\phi_k}&=\ln\left(1+\f{\phi_{k+1}-\phi_k}{\phi_k}\right)\\
&\ge\f{1}{2}\left(\f{\phi_{k+1}-\phi_k}{\phi_k}\right)\\
&\ge\f{1}{2\phi}\left(\phi_{k+1}-\phi_k\right).
\end{align*}
Using Propositions \ref{upper} and \ref{lower}, and setting $$A_k=\sqrt{5-k(k+2)\phi^{-2k-2}+2\phi^{-k}(k(\phi-3)-2)},$$ we have
\begin{align*}
\phi_{k+1}-\phi_k&\ge \phi\left(\f{\sqrt{5}\phi^{k+2}-(k+3)}{\sqrt{5}\phi^{k+2}-(k+2)}\right) -\f{-2+(k+1)(k+2)\phi^{-k}}{1+k(k+2)\phi^{-k-1}-A_k}\\
&=\phi\left(1-\f{1}{\sqrt{5}\phi^{k+2}-(k+2)}\right) +\f{2\phi^k-(k+1)(k+2)}{\phi^k+k(k+2)\phi^{-1}-A_k\phi^k}\\
&=-\f{1}{\sqrt{5}\phi^{k+2}-(k+2)} +\f{2\phi^k-(k+1)(k+2)+\phi^{k+1}+k(k+2)-A_k\phi^{k+1}}{\phi^k+k(k+2)\phi^{-1}-A_k\phi^k}\\
&=-\f{1}{\sqrt{5}\phi^{k+2}-(k+2)} +\f{\phi^{k+1}(\sqrt{5}-A_k)-(k+2)}{\phi^k+k(k+2)\phi^{-1}-A_k\phi^k}\\
&=\f{-\f{\phi^k+k(k+2)\phii-A_k\phi^k}{\sqrt{5}\phi^{k+2}-(k+2)} +\phi^{k+1}(\sqrt{5}-A_k)-(k+2)}{\phi^k+k(k+2)\phi^{-1}-A_k\phi^k}\\
&:=\f{-B_k+C_k}{\phi^k+k(k+2)\phi^{-1}-A_k\phi^k}\\
&=\f{B_k-C_k}{-\phi^k-k(k+2)\phi^{-1}+A_k\phi^k}.
\end{align*}
Let's investigate $B_k$ and $C_k$ closer by computing the limit of each as $k\to\infty$. We see
\begin{align*}
\lim_{k\to\infty}B_k&=\lim_{k\to\infty}\f{-A_k\phi^k+\phi^k+k(k+2)\phii}{\sqrt{5}\phi^{k+2}-(k+2)}\\
&=\f{1-\sqrt{5}}{\sqrt{5}\phi^2}.
\end{align*}
In addition,
\begin{align*}
\lim_{k\to\infty}C_k=&\lim_{k\to\infty}(\phi^{k+1}(\sqrt{5}-A_k)-(k+2))\\
=&\lim_{k\to\infty}\f{(\sqrt{5}\phi^{k+1}-(k+2))^2-(\phi^{k+1}A_k)^2}{\phi^{k+1}A_k+\sqrt{5}\phi^{k+1}-(k+2)}\\
=&\lim_{k\to\infty}\left(\f{5\phi^{2k+2}-2\phi^{k+1}\sqrt{5}(k+2)+(k+2)^2}{\phi^{k+1}A_k+\sqrt{5}\phi^{k+1}-(k+2)}\right.\\
&-\left.\f{\phi^{2k+2}(5-k(k+2)\phi^{-2k-2}+2\phi^{-k}(k\phi-3k-2))}{\phi^{k+1}A_k+\sqrt{5}\phi^{k+1}-(k+2)} \right)\\
=&\lim_{k\to\infty}\f{\phi^{k+1}(2-2\sqrt{5})+(k+2)(2k+2)}{\phi^{k+1}A_k+\sqrt{5}\phi^{k+1}-(k+2)}\\
=&\f{1-\sqrt{5}}{\sqrt{5}}
\end{align*}
So $\lim_{k\to\infty}(B_k-C_k)=\f{\sqrt{5}-1}{\phi\sqrt{5}}$. Therefore there exists a $K\in\mathbb N$ such that for all $k>K$, $B_k-C_k\ge\f{1}{2}\f{\sqrt{5}-1}{\phi\sqrt{5}}$.
Finally we return to estimating $\phi_{k+1}-\phi_k$ as follows:
\begin{align*}
\phi_{k+1}-\phi_k&\ge\f{\f{\sqrt{5}-1}{2\phi\sqrt{5}}}{-\phi^k-k(k+2)\phi^{-1}+A\phi^k}\\
&\ge\f{\f{\sqrt{5}-1}{2\phi\sqrt{5}}}{(\sqrt{5}-1)\phi^k}\\
&=\f{1}{2\sqrt{5}\phi^{k+1}}.
\end{align*}
All the pieces fit together to provide:
\begin{align*}
\f{\ln\phi_{k+1}}{\ln\f{\phi_{k+1}}{\phi_k}}&\le\f{\ln\phi_{k+1}}{\f{1}{2\phi}(\phi_{k+1}-\phi_k)}\\
&\le \f{\ln\phi_{k+1}}{\f{1}{2\phi}\cdot\f{1}{2\sqrt{5}\phi^{k+1}}}\\
&\le\phi^{k+2}4\sqrt{5}\ln\phi
\end{align*}
provided $k>K$.
\end{proof}
Note that if we had used (\ref{badest}) to obtain our upper bound on $\phi_k$, that would make $A_k=\sqrt{5}$ for all $k$ and $\lim_{k\to\infty} C_k$ would not exist.
\subsection{Proof of Theorem~\ref{upperbound}}
The previous proposition would be perfect for our purposes if it were not for the fact that the estimate only works beyond an unspecified term $K$. In fact, we know the lower bound used in the proof is only valid for $k>8$. Computer algebra systems can be used to verify that $K=12.$ However, Table \ref{minn} can be used for small $k$. Proposition \ref{condition} not only verifies that the growth of the terms in the second column of Table \ref{minn} are exponential but it also allows us to locate a splitting point, $k_s$, for the sum determining $G_n$ even for large $n$.
We are now ready to prove Theorem \ref{upperbound}
\begin{proof}
Let $k_s=\lfloor \log_\phi\f{n}{8\phi^2\sqrt{5}\ln\phi}\rfloor$. For all $12\le k\le k_s$, we have
\begin{align*}
\phi^{k+2}4\sqrt{5}\ln\phi\le&\phi^{k_s+2}4\sqrt{5}\ln\phi\\
\le&\f{n}{2}\\
\le&n-k_s \\
\le&n-k.
\end{align*}
Therefore by Proposition \ref{condition}, we have for all $12\le k\le k_s$,
\begin{equation}\label{big}\phi_k^{n-k}\le\phi_{k+1}^{n-k-1}. \end{equation}
Also by Table \ref{minn} we have for all $k<\min(12,k_s)$,
\begin{equation}\label{little}\phi_k^{n-k}\le\phi_{k+1}^{n-k-1}.\end{equation}
Putting (\ref{big}) and (\ref{little}) together we have for all $k\le k_s$,
\begin{equation}\label{nec}\phi_k^{n-k}\le\phi_{k+1}^{n-k-1}. \end{equation}
Now we split the sum determing $G_n$ and apply (\ref{nec}) to obtain
\begin{align*}
G_n&=\sum_{k=1}^{k_s} A_{n-k,k}+\sum_{k=k_s+1}^{n-1} A_{n-k,k}\\
&\le\sum_{k=1}^{k_s} A_{n-k,k}+\sum_{k=1}^{n-k_s-1} f_k\\
&\le\sum_{k=1}^{k_s} A_{n-k,k}+f_{n-k_s+1}\\
&\le K_1\sum_{k=1}^{k_s} \phi_k^{n-k}+f_{n-k_s+1}\\
&\le K_1k_s\phi^{n-k_s}+K_1\phi^{n-k_s+1}\\
&=K_1\phi^{n-k_s}(k_s+\phi)\\
&\le K_1\phi^{n-\log_\phi\f{n}{8\phi^2\sqrt{5}\ln\phi}}\left(\log_\phi\f{n}{8\phi^2\sqrt{5}\ln\phi}+1+\phi\right)\\
&\le K_18\phi^2\sqrt{5}\f{\phi^n}{n}(\ln n),
\end{align*}
where $K_1=\max_{k\le k_s}\{b_k|A_{m,k}