\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 0.7cm{\LARGE\bf Upper Bounds for Prime Gaps \\ \vskip .1in Related to Firoozbakht's Conjecture } \vskip 0.7cm \large Alexei Kourbatov\\ 15127 NE 24th St., \#578\\ Redmond, WA 98052 \\ USA\\ \href{mailto:akourbatov@gmail.com}{\tt akourbatov@gmail.com} \end{center} \vskip .2 in \begin{abstract}\noindent We study two kinds of conjectural bounds for the prime gap after the $k$th prime~$p_k$: (A) $p_{k+1} < p_k^{1+1/k}$ and (B) $p_{k+1}-p_k < \log^2 p_k - \log p_k - b$ for $k>9$. The upper bound (A) is equivalent to Firoozbakht's conjecture. We prove that (A) implies (B) with $b=1$; on the other hand, (B) with $b=1.17$ implies (A). We also give other sufficient conditions for (A) that have the form (B) with $b\to1$ as $k\to\infty$. \end{abstract} \section{Introduction} \label{one} In 1982 Firoozbakht proposed the following conjecture \cite[p.~185]{ribenboim}: \medskip\noindent {\bf Firoozbakht's Conjecture.} If $p_k$ is the $k$th prime, the sequence $(p_k^{1/k})_{k\in{\mathbb N}}$ is decreasing. \smallskip\noindent Equivalently, for all $k\ge1$, the prime $p_{k+1}$ is bounded by the inequality \begin{equation}\label{ineqf} p_{k+1} < p_k^{1+1/k}. \end{equation} Several authors \cite{rivera,sinha2010,zwsun2013,zwsun2012} have observed that \begin{itemize} \item Firoozbakht's conjecture (\ref{ineqf}) implies {\it Cram\'er's conjecture} $p_{k+1}-p_k=O(\log^2{\negthinspace}p_k)$ \cite{cramer}. \item If conjecture (\ref{ineqf}) is true and $k$ is large, then \begin{equation}\label{ineqln} p_{k+1}-p_k ~<~ \log^2 p_k - \log p_k. \end{equation} \end{itemize} In Section~\ref{two} we prove that (\ref{ineqf}) implies a sharper bound than (\ref{ineqln}): \begin{equation}\label{in2a} p_{k+1}-p_k ~<~ \log^2 p_k - \log p_k - b \quad\mbox { for all } k>9, \end{equation} with $b=1$. If the exact value of $k=\pi(p_k)$ is not available, then a violation of (\ref{ineqln}) or (\ref{in2a}) might be used to {\it disprove} Firoozbakht's conjecture (\ref{ineqf}). However, given a pair of primes $p_k$, $p_{k+1}$, the validity of (\ref{ineqln}) alone is not enough for the verification of (\ref{ineqf}). We discuss this in more detail in Section 3; see also \cite{kourbatov2015v}. In Section 4 we prove that (\ref{in2a}) with $b=1.17$ implies (\ref{ineqf}); we also give other sufficient conditions for (\ref{ineqf}). Probabilistic considerations \cite[OEIS \seqnum{A235402}]{cramer,kourbatov2014,kourbatov2015v} suggest that bounds (\ref{ineqf}), (\ref{ineqln}), (\ref{in2a}) hold for {\it almost all maximal gaps} between primes. \section{A corollary of Firoozbakht's conjecture} \label{two} \begin{theorem}\label{th1} If conjecture $(\ref{ineqf})$ is true, then $$ p_{k+1}-p_k ~<~ \log^2 p_k - \log p_k - 1 \quad \mbox { for all } k>9. $$ \end{theorem} \begin{proof} It is easy to check that \begin{equation}\label{in4} {x+\log^2 x \over \log x-1-{1\over\log x}} ~<~ {x\over\log x-1-{1\over\log x}-{1\over\log^2 x}} \ \ \mbox{ for }x\ge285967. \end{equation} Let $\pi(x)$ denote the prime-counting function. From Axler \cite[Corollary 3.6]{axler} we have \begin{equation}\label{in5} {x\over\log x-1-{1\over\log x}-{1\over\log^2 x}} ~<~ \pi(x) \quad\mbox{ for }x\ge1772201. \end{equation} Taking the log of both sides of (\ref{ineqf}) we find that Firoozbakht's conjecture (\ref{ineqf}) is equivalent to \begin{equation}\label{in6} k ~<~ {\log p_k \over \log p_{k+1}-\log p_k}. \end{equation} Let $k\ge133115$. Then $p_k\ge1772201$. By setting $x=p_k$ in (\ref{in4}) and (\ref{in5}), we see that inequalities (\ref{in4}), (\ref{in5}), (\ref{in6}) form a chain. Therefore, if Firoozbakht's conjecture is true, then \begin{equation}\label{in7} {p_k+\log^2 p_k \over \log p_k - 1 -{1\over\log p_k}} ~<~ {\log p_k \over \log p_{k+1}-\log p_k} \quad\mbox{ for }p_k\ge1772201. \end{equation} Cross-multiplying, we get \begin{equation}\label{in8} (\log p_{k+1}-\log p_k)(p_k+\log^2 p_k) ~<~ \log^2 p_k - \log p_k - 1. \end{equation} We have \begin{equation}\label{in9} {y \over x+y} ~<~ \log(x+y)-\log x \qquad\mbox{ for every } x,y>0. \end{equation} Setting $x=p_k$ and $y=p_{k+1}-p_k$, we can replace the left-hand side of (\ref{in8}) by a smaller quantity $(p_{k+1}-p_k)(p_k+\log^2 p_k)/p_{k+1}$ to obtain the inequality $$ {(p_{k+1}-p_k)(p_k+\log^2 p_k)\over p_{k+1}} ~<~ \log^2 p_k - \log p_k - 1, $$ which is equivalent to $$ (p_{k+1}-p_k)(p_k+\log^2 p_k) ~<~ (p_k + (p_{k+1}-p_k))(\log^2 p_k - \log p_k - 1), $$ $$ p_{k+1}-p_k ~<~ {p_k\over p_k+\log p_k + 1}(\log^2 p_k - \log p_k - 1). $$ This proves the theorem for every $k\ge133115$ because $p_k/(p_k+\log p_k + 1)<1$. Separately, for $99 \ \ (p_k\ge29), \end{equation} then Firoozbakht's conjecture $(\ref{ineqf})$ is true. \end{theorem} \begin{proof} From Axler \cite[Corollary 3.5]{axler} we have \begin{equation}\label{in11} \log x - 1 - {1.17\over\log x} ~<~ {x\over\pi(x)} \quad\mbox{ for every } x\ge5.43. \end{equation} Let $k>9$. Multiplying both sides of (\ref{in11}) by $\log x$, taking $x=p_k$, and using (\ref{in10}), we get \begin{equation}\label{in12} p_{k+1}-p_k ~<~ \log^2 p_k - \log p_k - 1.17 ~<~ {p_k\log p_k\over k} \quad\mbox{ for } p_k \ge 29; \end{equation} therefore, \begin{equation}\label{in13} {p_{k+1}-p_k \over p_k} ~<~ {\log p_k\over k} \qquad\mbox{ for } p_k\ge29. \end{equation} We have $$ \log(x+y)-\log x ~<~ {y \over x} \qquad\mbox{ for every } x,y>0. $$ Setting $x=p_k$ and $y=p_{k+1}-p_k$, we can replace the left-hand side of (\ref{in13}) by a smaller quantity $\log p_{k+1}-\log p_k$ to obtain the inequality \begin{equation}\label{in14} \log p_{k+1}-\log p_k ~<~ {\log p_k\over k}, \end{equation} which is equivalent to $$ \log_{p_k} {p_{k+1}\over p_k} ~<~ {1\over k}. $$ Now, exponentiation with base $p_k$ yields Firoozbakht's conjecture (\ref{ineqf}) for $p_k\ge29$. This completes the proof since for small $p_k$ conjecture (\ref{ineqf}) holds unconditionally \cite{kourbatov2015v}. \end{proof} \smallskip\noindent {\bf Other sufficient conditions for (\ref{ineqf}). } Based on the $\pi(x)$ formula of Panaitopol \cite{panaitopol}, Axler gives a family of upper bounds for $\pi(x)$ \cite[Corollary 3.5]{axler}: \begin{eqnarray*} \pi(x) &<& {x\over\log x - 1 - {1.17\over\log x}} \mspace{190mu} \mbox{ for } x\ge5.43, \\ \pi(x) &<& {x\over\log x - 1 - {1\over\log x} - {3.83\over\log^2 x}} \mspace{133mu} \mbox{ for } x\ge9.25, \\ \pi(x) &<& {x\over\log x - 1 - {1\over\log x} - {3.35\over\log^2 x} - {15.43\over\log^3 x}} \mspace{75mu} \mbox{ for } x\ge14.36, \\ \pi(x) &<& {x\over\log x - 1 - {1\over\log x} - {3.35\over\log^2 x} - {12.65\over\log^3 x} - {89.6\over\log^4 x}} \quad \mbox{ for } x\ge21.95. \\ \end{eqnarray*} Just as in Theorem \ref{th3}, we can transform the above upper bounds into sufficient conditions for Firoozbakht's conjecture (\ref{ineqf}) and obtain our next theorem. \begin{theorem}\label{th4} If one or more of the following conditions hold for all $p_k>4\times10^{18}:$ \begin{eqnarray*} p_{k+1}-p_k &<& \log^2 p_k - \log p_k - 1.17, \\ p_{k+1}-p_k &<& \log^2 p_k - \log p_k - 1 - {3.83\over\log p_k}, \phantom{1^{1^{1^1}}\over1} \\ p_{k+1}-p_k &<& \log^2 p_k - \log p_k - 1 - {3.35\over\log p_k} - {15.43\over\log^2 p_k}, \\ p_{k+1}-p_k &<& \log^2 p_k - \log p_k - 1 - {3.35\over\log p_k} - {12.65\over\log^2 p_k} - {89.6\over\log^3 p_k}, \end{eqnarray*} then Firoozbakht's conjecture $(\ref{ineqf})$ is true. \end{theorem} In the statement of Theorem \ref{th4}, we have taken into account that for $p_k<4\times10^{18}$ conjecture $(\ref{ineqf})$ holds unconditionally \cite{kourbatov2015v}. We do not give a proof of Theorem \ref{th4}; it is fully similar to the proof of Theorem \ref{th3}. \medskip\noindent {\bf Remarks.} (i) In inequality (\ref{in10}) the right-hand side is an increasing function of $p_k$. Therefore, if (\ref{in10}) holds for a {\it maximal prime gap} with $p_k = \mbox{\seqnum{A002386}}(n)$, then (\ref{in10}) must also be true for {\em every} $p_k$ between $\mbox{\seqnum{A002386}}(n)$ and $\mbox{\seqnum{A002386}}(n+1)$. So an easy way to prove Theorem \ref{th2} is to check (\ref{ineqf}) directly for all primes $p_k\le89$, then verify (\ref{in10}) just for maximal prime gaps with $p_k=\mbox{\seqnum{A002386}}(n)\ge89$. (ii) In Theorem \ref{th4}, the coefficients of $(\log p_k)^{-n}$ approximate the terms of OEIS sequence \seqnum{A233824}: a recurrent sequence in Panaitopol's formula for $\pi(x)$ \cite{panaitopol}. \section{Appendix: An asymptotic formula for $p_k^{1+1/k}-p_k$} \label{five} \begin{theorem}\label{th5} Let $p_k$ be the $k$-th prime, and let $f_k = p_k^{1+1/k}- p_k$, then $$ f_k = \log^2 p_k - \log p_k - 1 +o(1) \quad\mbox{ as } k\to\infty \qquad\mbox{\rm(cf.~OEIS \seqnum{A246778}).} $$ \end{theorem} \begin{proof} From Axler \cite[Corollaries 3.5, 3.6]{axler} we have \begin{equation}\label{dblin} {x\over\log x-1-{1\over\log x}-{1\over\log^2 x}} ~<~ \pi(x) ~<~ {x\over\log x-1-{1\over\log x}-{3.83\over\log^2 x}} \quad\mbox{ for } x\ge1772201. \end{equation} By definition of $f_k$, we have $\log_{p_k}(p_k+f_k) = 1+1/k$, so $\displaystyle k = \pi(p_k) = {\log p_k \over \log(p_k+f_k)-\log p_k}. $ Therefore, for $x=p_k\ge1772201$, we can rewrite (\ref{dblin}) as \begin{equation}\label{dblin2} {p_k\over\log p_k-1-{1\over\log p_k}-{1\over\log^2 p_k}} ~<~ {\log p_k \over \log(p_k+f_k)-\log p_k} ~<~ {p_k\over\log p_k-1-{1\over\log p_k}-{3.83\over\log^2 p_k}}. \end{equation} {\em An upper bound for} $f_k$. We combine (\ref{in4}) with the left inequality of (\ref{dblin2}) to get \begin{equation}\label{inA} {p_k+\log^2 p_k \over \log p_k - 1 -{1\over\log p_k}} ~<~ {\log p_k \over \log(p_k + f_k)-\log p_k} \quad\mbox{ for }p_k\ge1772201. \end{equation} Cross-multiplying and using (\ref{in9}), similar to Theorem \ref{th1}, we obtain $$ {f_k(p_k+\log^2 p_k)\over p_k + f_k} ~<~ {(\log(p_k + f_k)-\log p_k)(p_k+\log^2 p_k)} ~<~ \log^2 p_k - \log p_k - 1, $$ $$ f_k(p_k+\log^2 p_k) ~<~ (p_k + f_k)(\log^2 p_k - \log p_k - 1), $$ $$ f_k ~<~ {p_k\over p_k+\log p_k + 1}(\log^2 p_k - \log p_k - 1) ~<~ \log^2 p_k - \log p_k - 1. $$ {\em A lower bound for} $f_k$. From the right inequality of (\ref{dblin2}) we get $$ {\log^2 p_k - \log p_k - 1 - {3.83\over\log p_k}\over p_k} ~<~ \log(p_k + f_k)-\log p_k ~<~ {f_k\over p_k}, $$ $$ \log^2 p_k - \log p_k - 1 - {3.83\over\log p_k} ~<~ f_k. $$ Together, the upper and lower bounds yield the desired asymptotic formula for $k\to\infty$. \end{proof} \section{Acknowledgments} \label{six} The author expresses his gratitude to the anonymous referee for numerous useful suggestions, and to all contributors and editors of the websites {\tt OEIS.org} and {\tt PrimePuzzles.net}, particularly to Farideh Firoozbakht for proposing a very interesting conjecture. Thanks are also due to Christian Axler for proving the $\pi(x)$ bounds \cite{axler} used in Theorems \ref{th1}, \ref{th3}--\ref{th5}. \begin{thebibliography}{20} \bibitem{axler} C.~Axler, New bounds for the prime counting function $\pi(x)$, preprint, 2014, \\ \url{http://arxiv.org/abs/1409.1780}. \bibitem{cramer} H.~Cram\'er, On the order of magnitude of the difference between consecutive prime numbers, {\it Acta Arith.} {\bf 2} (1936), 23--46. \bibitem{kourbatov2014} A.~Kourbatov, The distribution of maximal prime gaps in Cram\'er's probabilistic model of primes, {\it Int.~J. Statistics and Probability} {\bf 3} (2014), 18--29. \bibitem{kourbatov2015v} A.~Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, {\it Int.~Math.~Forum} {\bf 10} (2015), 283--288. \bibitem{panaitopol} L.~Panaitopol, A formula for $\pi(x)$ applied to a result of Koninck-Ivi\'c, {\it Nieuw Arch. Wiskd.} {\bf 5} (2000), 55--56. \bibitem{ribenboim} P.~Ribenboim, {\it The Little Book of Bigger Primes}, New York, Springer, 2004. \bibitem{rivera} C. Rivera, ed., Conjecture 30: the Firoozbakht conjecture, 2002. \\ Available at \url{http://www.primepuzzles.net/conjectures/conj_030.htm}. \bibitem{sinha2010} N.~K.~Sinha, On a new property of primes that leads to a generalization of Cram\'er's conjecture, preprint, 2010, \url{http://arxiv.org/abs/1010.1399}. \bibitem{oeis} N.~J.~A.~Sloane, ed., {\it The On-Line Encyclopedia of Integer Sequences}, 2015. Published electronically at \url{http://oeis.org/A111943}. \bibitem{zwsun2013} Z.-W. Sun, On a sequence involving sums of primes, {\em Bull.~Aust.~Math.~Soc.} {\bf 88} (2013), 197--205. \bibitem{zwsun2012} Z.-W. Sun, Various new observations about primes, talk given at Univ.\ of Illinois at Urbana-Champaign, August 28 2012. Available at \\ \url{http://math.nju.edu.cn/~zwsun/ObservPrime.pdf}. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: 11N05. \noindent \emph{Keywords: } Cram\'er conjecture, Firoozbakht conjecture, prime gap. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A002386}, \seqnum{A005250}, \seqnum{A111943}, \seqnum{A182134}, \seqnum{A182514}, \seqnum{A182519}, \seqnum{A205827}, \seqnum{A233824}, \seqnum{A235402}, \seqnum{A235492}, \seqnum{A245396}, \seqnum{A246776}, \seqnum{A246777}, \seqnum{A246778}, \seqnum{A246810}, \seqnum{A249669}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received June 18 2015; revised versions received September 26 2015; October 1 2015. Published in {\it Journal of Integer Sequences}, November 24 2015. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}