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

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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







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  • 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






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share|cite|improve this question













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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






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oldest

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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.






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    1 Answer
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    active

    oldest

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    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




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