Is every prime that is greater than 5, less than the sum of the two previous primes?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite












Can one prove by elementary means (such as Paul Erdös' proof of Bertrand's Postulate) that every prime greater than 5 is less than the sum of the two primes immediately preceding it?






nt.number-theory prime-numbers







share|cite|improve this question



















  • 4




    From this, $p_n+2leq frac32p_n+1leq p_n+1+frac34p_n$.
    – Wojowu
    3 hours ago






  • 2




    Prof. Bernardo Recamán Santos: the answer is YES, essentially this question was the subject matter of this previous question in MO: mathoverflow.net/q/113840/1593
    – José Hdz. Stgo.
    3 hours ago







  • 2




    @Wojowu: El Bachraoui's result was already established by Schur in 1929, see Jose Brox's answer here: mathoverflow.net/questions/289402/… .
    – Ofir Gorodetsky
    3 hours ago















up vote
2
down vote

favorite












Can one prove by elementary means (such as Paul Erdös' proof of Bertrand's Postulate) that every prime greater than 5 is less than the sum of the two primes immediately preceding it?






nt.number-theory prime-numbers







share|cite|improve this question



















  • 4




    From this, $p_n+2leq frac32p_n+1leq p_n+1+frac34p_n$.
    – Wojowu
    3 hours ago






  • 2




    Prof. Bernardo Recamán Santos: the answer is YES, essentially this question was the subject matter of this previous question in MO: mathoverflow.net/q/113840/1593
    – José Hdz. Stgo.
    3 hours ago







  • 2




    @Wojowu: El Bachraoui's result was already established by Schur in 1929, see Jose Brox's answer here: mathoverflow.net/questions/289402/… .
    – Ofir Gorodetsky
    3 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Can one prove by elementary means (such as Paul Erdös' proof of Bertrand's Postulate) that every prime greater than 5 is less than the sum of the two primes immediately preceding it?






nt.number-theory prime-numbers







share|cite|improve this question















Can one prove by elementary means (such as Paul Erdös' proof of Bertrand's Postulate) that every prime greater than 5 is less than the sum of the two primes immediately preceding it?






nt.number-theory prime-numbers




nt.number-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









David Roberts

16.2k461170




16.2k461170










asked 3 hours ago









Bernardo Recamán Santos

823526




823526







  • 4




    From this, $p_n+2leq frac32p_n+1leq p_n+1+frac34p_n$.
    – Wojowu
    3 hours ago






  • 2




    Prof. Bernardo Recamán Santos: the answer is YES, essentially this question was the subject matter of this previous question in MO: mathoverflow.net/q/113840/1593
    – José Hdz. Stgo.
    3 hours ago







  • 2




    @Wojowu: El Bachraoui's result was already established by Schur in 1929, see Jose Brox's answer here: mathoverflow.net/questions/289402/… .
    – Ofir Gorodetsky
    3 hours ago













  • 4




    From this, $p_n+2leq frac32p_n+1leq p_n+1+frac34p_n$.
    – Wojowu
    3 hours ago






  • 2




    Prof. Bernardo Recamán Santos: the answer is YES, essentially this question was the subject matter of this previous question in MO: mathoverflow.net/q/113840/1593
    – José Hdz. Stgo.
    3 hours ago







  • 2




    @Wojowu: El Bachraoui's result was already established by Schur in 1929, see Jose Brox's answer here: mathoverflow.net/questions/289402/… .
    – Ofir Gorodetsky
    3 hours ago








4




4




From this, $p_n+2leq frac32p_n+1leq p_n+1+frac34p_n$.
– Wojowu
3 hours ago




From this, $p_n+2leq frac32p_n+1leq p_n+1+frac34p_n$.
– Wojowu
3 hours ago




2




2




Prof. Bernardo Recamán Santos: the answer is YES, essentially this question was the subject matter of this previous question in MO: mathoverflow.net/q/113840/1593
– José Hdz. Stgo.
3 hours ago





Prof. Bernardo Recamán Santos: the answer is YES, essentially this question was the subject matter of this previous question in MO: mathoverflow.net/q/113840/1593
– José Hdz. Stgo.
3 hours ago





2




2




@Wojowu: El Bachraoui's result was already established by Schur in 1929, see Jose Brox's answer here: mathoverflow.net/questions/289402/… .
– Ofir Gorodetsky
3 hours ago





@Wojowu: El Bachraoui's result was already established by Schur in 1929, see Jose Brox's answer here: mathoverflow.net/questions/289402/… .
– Ofir Gorodetsky
3 hours ago











1 Answer
1






active

oldest

votes

















up vote
5
down vote



accepted










I am to elaborate a bit on how it is that the answer I left in this previous question implies a positive answer to Mr. Recamán's question.



By elementary means (i.e., without resorting to any complex analysis), in his paper Mémoire sur les nombres premiers (Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.), P. L. Chebyshev established that for any $varepsilon >frac15$, there exists an $n_varepsilonin mathbbN$ such that for all $ngeq n_varepsilon$,



$$pi((1+varepsilon)n)−pi(n)>0.$$



In particular, taking $varepsilon = frac14$, this implies that, for every sufficiently large $k in mathbbN$, we have that



$$p_k+2 leq (1.25)^2p_k < 2p_k$$



and whence, for every sufficiently large $k in mathbbN$, it is certainly the case that



$$p_k+2 < 2p_k < p_k+p_k+1.$$



By retracing Chebyshev's arguments it is possible to obtain an explicit version of the previous conclusion.






share|cite|improve this answer






















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "504"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f312509%2fis-every-prime-that-is-greater-than-5-less-than-the-sum-of-the-two-previous-pri%23new-answer', 'question_page');

    );

    Post as a guest

















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    I am to elaborate a bit on how it is that the answer I left in this previous question implies a positive answer to Mr. Recamán's question.



    By elementary means (i.e., without resorting to any complex analysis), in his paper Mémoire sur les nombres premiers (Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.), P. L. Chebyshev established that for any $varepsilon >frac15$, there exists an $n_varepsilonin mathbbN$ such that for all $ngeq n_varepsilon$,



    $$pi((1+varepsilon)n)−pi(n)>0.$$



    In particular, taking $varepsilon = frac14$, this implies that, for every sufficiently large $k in mathbbN$, we have that



    $$p_k+2 leq (1.25)^2p_k < 2p_k$$



    and whence, for every sufficiently large $k in mathbbN$, it is certainly the case that



    $$p_k+2 < 2p_k < p_k+p_k+1.$$



    By retracing Chebyshev's arguments it is possible to obtain an explicit version of the previous conclusion.






    share|cite|improve this answer


























      up vote
      5
      down vote



      accepted










      I am to elaborate a bit on how it is that the answer I left in this previous question implies a positive answer to Mr. Recamán's question.



      By elementary means (i.e., without resorting to any complex analysis), in his paper Mémoire sur les nombres premiers (Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.), P. L. Chebyshev established that for any $varepsilon >frac15$, there exists an $n_varepsilonin mathbbN$ such that for all $ngeq n_varepsilon$,



      $$pi((1+varepsilon)n)−pi(n)>0.$$



      In particular, taking $varepsilon = frac14$, this implies that, for every sufficiently large $k in mathbbN$, we have that



      $$p_k+2 leq (1.25)^2p_k < 2p_k$$



      and whence, for every sufficiently large $k in mathbbN$, it is certainly the case that



      $$p_k+2 < 2p_k < p_k+p_k+1.$$



      By retracing Chebyshev's arguments it is possible to obtain an explicit version of the previous conclusion.






      share|cite|improve this answer
























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        I am to elaborate a bit on how it is that the answer I left in this previous question implies a positive answer to Mr. Recamán's question.



        By elementary means (i.e., without resorting to any complex analysis), in his paper Mémoire sur les nombres premiers (Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.), P. L. Chebyshev established that for any $varepsilon >frac15$, there exists an $n_varepsilonin mathbbN$ such that for all $ngeq n_varepsilon$,



        $$pi((1+varepsilon)n)−pi(n)>0.$$



        In particular, taking $varepsilon = frac14$, this implies that, for every sufficiently large $k in mathbbN$, we have that



        $$p_k+2 leq (1.25)^2p_k < 2p_k$$



        and whence, for every sufficiently large $k in mathbbN$, it is certainly the case that



        $$p_k+2 < 2p_k < p_k+p_k+1.$$



        By retracing Chebyshev's arguments it is possible to obtain an explicit version of the previous conclusion.






        share|cite|improve this answer














        I am to elaborate a bit on how it is that the answer I left in this previous question implies a positive answer to Mr. Recamán's question.



        By elementary means (i.e., without resorting to any complex analysis), in his paper Mémoire sur les nombres premiers (Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.), P. L. Chebyshev established that for any $varepsilon >frac15$, there exists an $n_varepsilonin mathbbN$ such that for all $ngeq n_varepsilon$,



        $$pi((1+varepsilon)n)−pi(n)>0.$$



        In particular, taking $varepsilon = frac14$, this implies that, for every sufficiently large $k in mathbbN$, we have that



        $$p_k+2 leq (1.25)^2p_k < 2p_k$$



        and whence, for every sufficiently large $k in mathbbN$, it is certainly the case that



        $$p_k+2 < 2p_k < p_k+p_k+1.$$



        By retracing Chebyshev's arguments it is possible to obtain an explicit version of the previous conclusion.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 hours ago

























        answered 2 hours ago









        José Hdz. Stgo.

        5,03134674




        5,03134674



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f312509%2fis-every-prime-that-is-greater-than-5-less-than-the-sum-of-the-two-previous-pri%23new-answer', 'question_page');

            );

            Post as a guest
































































            Comments

            Post a Comment

            Popular posts from this blog

            What does second last employer means? [closed]

            Image
            Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
            Read more

            Installing NextGIS Connect into QGIS 3?

            Image
            Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
            Read more

            One-line joke

            A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
            Read more