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\title{Paper template}
\author{Your Name}
\email{\textcolor{blue}{\href{mailto:youremail@school.edu}{youremail@school.edu}}}
\address{Department of Mathematics, School, City, Zip}
\author{Steven J. Miller}
\email{\textcolor{blue}{\href{mailto:sjm1@williams.edu}{sjm1@williams.edu}}, \textcolor{blue}{\href{Steven.Miller.MC.96@aya.yale.edu}{Steven.Miller.MC.96@aya.yale.edu}}}
\address{Department of Mathematics and Statistics, Williams College, Williamstown, MA 01267}
\thanks{Miller was partially supported by NSF grant DMS1265673. We thank....}
\subjclass[2010]{60B10, 11B39, 11B05 (primary) 65Q30 (secondary)}
\keywords{Zeckendorf decompositions, Fibonacci quilt, non-uniqueness of representations, positive linear recurrence relations, Gaussian behavior, distribution of gaps}
\date{\today}
\begin{document}
\maketitle
\begin{abstract} Zeckendorf's theorem states something.
\end{abstract}
\tableofcontents
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\section{Introduction}
%\fix{SM: My feelings are we should emphasize how we develop a new technique that allows us to bypass the technical obstructions that have plagued much of the previous literature, specifically having to do a lot of work with quantities related to general characteristic polynomials, and it's a pain to analyze these in general. We're able to prove the coefficients of the linear terms in the mean and variance are non-zero elementarily, bypassing the need for such arguments. Next major contribution is continuing our recent study of cases where the leading term of the recurrence is zero. Here we find some very different behavior (Fib Quilt), as well as similar behavior ($(s,b)$). In the interest of space we will say that we are omitting many standard calculations, referring the reader to either the appendix to the arXiv version or the companion CANT paper (I have a decent B-level journal in mind for this paper, have refereed for them recently, including in fact a paper that could not deal with the obstructions caused by general polynomials, and thus there is a chance they would be open to the CANT paper). I have NOT looked at the introduction for now.}
%\fix{SM: Rephrase intro to highlight connection between $(s,b)$ and FibQuilt.}
Previous work on ....
\begin{defi}[($s,b$)-Generacci sequence]\label{sbDefi}
For fixed integers $s, b \geq 1$, an increasing sequence of positive integers $\{a_i\}_{i=1}^\infty$ is the {\em $(s,b)$-Generacci sequence} if every $a_i$ for $i \geq 1$ is the smallest positive integer that does not have an $(s,b)$-Generacci legal decomposition using the elements $\{a_1, \dots, a_{i-1}\}$.
\end{defi}
We recall that Zeckendorf's theorem gave an equivalent definition of the Fibonacci numbers as the unique sequence which allows one to write all positive integers as a sum of nonconsecutive elements in the sequence. Note this holds provided we define the Fibonacci numbers beginning with $1,2,3,\ldots$. It is then clear that the $(1,1)$-Generacci sequence is the Fibonacci sequence. However, other interesting sequences are also $(s,b)$-Generacci sequences. For example, Narayana's cow sequence is the $(2,1)$-Generacci sequence and the Kentucky sequence (studied at length by the authors in \cite{LiM}) is the $(1,2)$-Generacci sequence.
\begin{thm}[Recurrence Relation and Explicit Formula]\label{thrm:recurrencesb_2}
For $n > (s+1)b+1$, the $n^{\text{th}}$ term of the $(s,b)$-Generacci sequence satisfies
\begin{equation}\label{explicitConstant_2}
a_n \ = \ a_{n-b}+ba_{n-(s+1)b} \ = \ c_1\lambda_1^n \left[1 + O\left((\lambda_2/\lambda_1)^n\right)\right],
\end{equation}
where $\lambda_1$ is the largest root of $x^{(s+1)b} - x^{sb} - b = 0$, and $c_1$ and $\lambda_2$ are constants with $\lambda_1>1$, $c_1 > 0 $ and $|\lambda_2| < \lambda_1$.
\end{thm}
We thus see something....
\begin{rek} Here is a remark. This makes the text stand out.... \end{rek}
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\subsubsection{Fibonacci Quilt Sequence} Blah blah.
\begin{thm}[Recurrence Relations]\label{thm:rrfibquilt}
Let $q_n$ denote the $n$\textsuperscript{th} term in the Fibonacci Quilt Sequence. Then
(1) for $n\geq 6$, $q_{n+1} = q_{n} + q_{n-4}$, (2) for $n \geq 5$, $q_{n+1} = q_{n-1} + q_{n-2}$, and (3) we have \be q_n \ = \ \alpha_1 \lambda_1^n + \alpha_2 \lambda_2^n + \alpha_3 \overline{\lambda_2}^n, \ee where $\alpha_1 \approx 1.26724 $, \be \lambda_1 \ = \ \frac13 \left(\frac{27}{2} - \frac{3 \sqrt{69}}{2}\right)^{1/3} + \frac{\left(\frac12 \left(9 + \sqrt{69}\right)\right)^{1/3}}{3^{2/3}} \ \approx \ 1.32472 \ee and $\lambda_2 \approx -0.662359 - 0.56228i$ (which has absolute value approximately 0.8688).
\end{thm}
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\subsection{Results}
Both the $(s,b)$-Generacci sequences and the Fibonacci quilt sequence satisfy recurrence relations with leading term zero. They display drastically different behavior in some respects, but also have very similar behavior for other problems (which allows us to deduce results for the Fibonacci Quilt sequence from results for the $(4,1)$-Generacci sequence). We begin by stating results related to the decompositions arising from these sequences, many of which are proved in the companion paper \cite{CFHMN2}. We then state new results on Gaussian behavior in the number of summands, and exponential decay in the gap measures between summands.
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\subsubsection{Decompositions}
The $(s,b)$-Generacci legal decompositions are unique.
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\subsubsection{Wickets}
Never have a subsubsection unless you have at least two.
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\subsection{Comments}
Similar to the above, never have a subsection unless you have at least two.
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\section{Gap Measures}
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\subsection{Average Bin Gap Measure}\label{sec:gaps}
\begin{proof}[Proof of Theorem ADD REF]
Let $m \in I_n:= [a_{(n-1)b+1},a_{nb+1})$ have a legal decomposition.
Thus this case contributes to $H_{n+1}$ \bea & & (H_{n-1} - H_{n-Z}) + (c_2-1)(H_{n-1} - H_{n-1-Z}) \nonumber\\ & & \ \ \ \ \ \ = \ c_2 (H_{n-1} - H_{n-1-Z}) + (H_{n-1-Z}-H_{n-Z}). \eea This completes the proof....
\end{proof}
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\subsection{Individual Bin Gap Measure}
Now do it for an individual....
\appendix
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\section{Computations}
\begin{proof} For any integer $N$ let $p_N$ and $q_N$ be the relatively prime integers satisfying \be \frac{p_N}{q_N} \ = \ \prod_{p \le N} \left(1 - \frac1{p^2}\right)^{-1} \ = \ \prod_{p \le N} \left(1 + \frac{1}{p^2 - 1}\right). \ \ee \emph{Assume there are no primes $p \in (N,f(N)]$, where $f(x)$ is some rapidly growing function to be determined later.} If $f(N)$ is too large relative to $N$, we will find that $p_N/q_N$ is too good of a rational approximation to $\pi^2/6$, and thus there must be at least one prime between $N$ and $f(N)$. Under our assumption, we find \be \left|\frac{p_N}{q_N} - \frac{\pi^2}6\right| \ = \ \frac{p_N}{q_N} \left|1 - \prod_{p > f(N)} \left(1 + \frac{1}{p^2 - 1}\right)\right|. \ \ee Clearly $p_N/q_N \le \pi^2/6$, and \bea \prod_{p > f(N)} \left(1 + \frac{1}{p^2 - 1}\right) & \ = \ & \exp\left(\log \prod_{p > f(N)} \left(1 + \frac{1}{p^2 - 1}\right) \right) \nonumber\\ &\le & \exp\left( \sum_{n > f(N)} \log\left(1 + \frac1{(n-1)^2}\right) \right) \\ & \le & \exp\left( \sum_{n > f(N)} \frac1{(n-1)^2}\right) \ \le \ \exp\left( \frac{1}{f(N)^2} + \frac{1}{f(N)}\right) \ \eea (the last inequality follows by the replacing the sum over $n \ge f(N)+2$ with an integral). Standard properties of the exponential function yield \be \left|\frac{p_N}{q_N} - \frac{\pi^2}6\right| \ \le \ \frac{\pi^2}{6} \left|1 - \exp\left( \frac{1}{f(N)^2} + \frac{1}{f(N)}\right)\right| \ \le \ \frac{10}{f(N)}. \ \ee The largest $q_N$ can be is $N!^2$, which happens only if all integers at most $N$ are prime. Obviously we can greatly reduce this bound, as the only even prime is $2$; however, our purpose is to highlight the method by using the most elementary arguments possible. If we take $f(x) = (x!)^{14}$, we find (for $N$ sufficiently large) that \be\label{eq:temppisquareddenom} \left| \frac{\pi^2}6 - \frac{p_N}{q_N}\right| \ \le \ \frac{10}{f(N)} \ < \ \frac{1}{q_N^6}; \ee however, this contradicts Rhin and Viola's bound on the irrationality measure of $\pi^2/6$ ($\mu_{\rm irr}(\pi^2/6)$ $<$ $5.45$). Thus there must be a prime between $N$ and $f(N)$. Note $f(N) \le N^{14N} \le (14N)^{14N}$ for large $N$. Letting $f^{(k)}(N)$ denote the result of applying $f$ a total of $k$ times to $N$, for $N_0$ sufficiently large we see for large $k$ that there are at least $k$ primes at most $T(14 N_0,2k)$.
\end{proof}
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\begin{thebibliography}{BBBGILMT1} % '2nd argument contains the widest acronym'
\bibitem[BBGILMT]{BBGILMT}
O. Beckwith, A. Bower, L. Gaudet, R. Insoft, S. Li, S. J. Miller and P. Tosteson, \emph{The Average Gap Distribution for Generalized Zeckendorf Decompositions}, Fibonacci Quarterly \textbf{51} (2013), 13--27.
\bibitem[LiM]{LiM}
R. Li and S. J. Miller, \emph{Central Limit Theorems for gaps of Generalized Zeckendorf Decompositions}, preprint 2016. \bburl{http://arxiv.org/pdf/1606.08110}.
\bibitem[Ze]{Ze} E. Zeckendorf, \emph{Repr\'esentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas}, Bulletin de la Soci\'et\'e Royale des Sciences de Li\'ege \textbf{41} (1972), pages 179--182.
\end{thebibliography}
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\end{document}