Using least common multiple to prove there exists a prime between $2x$ and $3x$












2












$begingroup$


Let $text{lcm}(x)$ be the least common multiple of ${1,2,3,dots, x}$.



Hanson showed that $text{lcm}(x) < 3^x$



I'm wondering if the following argument is valid for showing that there is always a prime $p$ such that $2x < p < 3x$ for $x ge 246$.



Here is my argument:



(1) if prime $p$ satisfies $2x < p < 3x$, then $p | {{3x}choose{2x}}$.




This follows from the fact that ${{3x}choose{2x}}$ is an integer and that $p$ will not divide out by $2x!$ or $x!$.




(2) ${{3x}choose{2x}} > dfrac{6^x}{x}$ for $x ge 4$




For $x ge 4$, ${{2x}choose{x}} ge dfrac{4^x}{x}$ since ${8choose4} = 70 > dfrac{4^4}{4} = 64$ and ${{2x}choose{x}} = 2left(dfrac{2x-1}{x}right){2(x-1)choose{x-1}} > 2left(dfrac{2x-1}{x}right)left(dfrac{4^{x-1}}{x-1}right) > dfrac{4^x}{x}$ and ${{3x}choose{2x}} > left(dfrac{3x}{2x}right)^x {{2x}choose{x}} > left(dfrac{3}{2}right)^xdfrac{4^x}{x}$




(3) For any prime $q$ such that $frac{3x}{2} le q le 2x$, it follows that $2q > 3x$ and $q nmid {{3x}choose{2x}}$ since it will be divided out by $2x!$.



(4) Assume that there is no prime greater than $2x$ that divides ${{3x}choose{2x}}$



(5) It follows that ${{3x}choose{2x}} < text{lcm}(frac{3x}{2})text{lcm}(sqrt{3x})$ since:





  • From Legendre's Formula, it is well known that if $v_p(x)$ is the highest power of $p$ that divides $x$, then $v_p({{3x}choose{2x}}) = sumlimits_{i le log_p(3n)} leftlfloorfrac{3x}{p^i}rightrfloor - leftlfloorfrac{2x}{p^i}rightrfloor - leftlfloorfrac{x}{p^i}rightrfloor$ which equals $1$ or $0$ for each $i$ so that $v_p({{3x}choose{2x}}) le log_p(3x)$


  • If a prime $p > sqrt{3x}$, then $v_p({{3x}choose{2x}}) = 1$ and $p | text{lcm}(frac{3x}{2})$


  • If a prime $p le sqrt{3x}$, then $v_p({{3x}choose{2x}}) le log_p(3x) le v_p(frac{3x}{2})+1$





(6) So that: $dfrac{6^x}{x} < 3^{3x/2}3^{sqrt{3x}}$



(7) But this is not true for $x ge 246$ since:




$246ln 6 - ln 246 > 435.26 > 435.24 > frac{3}{2}(246)ln 3 + sqrt{3times246}ln 3$



and for $x ge 246$, $6 > (3^{3/2})(3^{sqrt{3x+2} - sqrt{3x}})left(dfrac{x+1}{x}right)$




(8) So we can reject step (4).






elementary-number-theory proof-verification prime-numbers least-common-multiple







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $text{lcm}(x)$ be the least common multiple of ${1,2,3,dots, x}$.



    Hanson showed that $text{lcm}(x) < 3^x$



    I'm wondering if the following argument is valid for showing that there is always a prime $p$ such that $2x < p < 3x$ for $x ge 246$.



    Here is my argument:



    (1) if prime $p$ satisfies $2x < p < 3x$, then $p | {{3x}choose{2x}}$.




    This follows from the fact that ${{3x}choose{2x}}$ is an integer and that $p$ will not divide out by $2x!$ or $x!$.




    (2) ${{3x}choose{2x}} > dfrac{6^x}{x}$ for $x ge 4$




    For $x ge 4$, ${{2x}choose{x}} ge dfrac{4^x}{x}$ since ${8choose4} = 70 > dfrac{4^4}{4} = 64$ and ${{2x}choose{x}} = 2left(dfrac{2x-1}{x}right){2(x-1)choose{x-1}} > 2left(dfrac{2x-1}{x}right)left(dfrac{4^{x-1}}{x-1}right) > dfrac{4^x}{x}$ and ${{3x}choose{2x}} > left(dfrac{3x}{2x}right)^x {{2x}choose{x}} > left(dfrac{3}{2}right)^xdfrac{4^x}{x}$




    (3) For any prime $q$ such that $frac{3x}{2} le q le 2x$, it follows that $2q > 3x$ and $q nmid {{3x}choose{2x}}$ since it will be divided out by $2x!$.



    (4) Assume that there is no prime greater than $2x$ that divides ${{3x}choose{2x}}$



    (5) It follows that ${{3x}choose{2x}} < text{lcm}(frac{3x}{2})text{lcm}(sqrt{3x})$ since:





    • From Legendre's Formula, it is well known that if $v_p(x)$ is the highest power of $p$ that divides $x$, then $v_p({{3x}choose{2x}}) = sumlimits_{i le log_p(3n)} leftlfloorfrac{3x}{p^i}rightrfloor - leftlfloorfrac{2x}{p^i}rightrfloor - leftlfloorfrac{x}{p^i}rightrfloor$ which equals $1$ or $0$ for each $i$ so that $v_p({{3x}choose{2x}}) le log_p(3x)$


    • If a prime $p > sqrt{3x}$, then $v_p({{3x}choose{2x}}) = 1$ and $p | text{lcm}(frac{3x}{2})$


    • If a prime $p le sqrt{3x}$, then $v_p({{3x}choose{2x}}) le log_p(3x) le v_p(frac{3x}{2})+1$





    (6) So that: $dfrac{6^x}{x} < 3^{3x/2}3^{sqrt{3x}}$



    (7) But this is not true for $x ge 246$ since:




    $246ln 6 - ln 246 > 435.26 > 435.24 > frac{3}{2}(246)ln 3 + sqrt{3times246}ln 3$



    and for $x ge 246$, $6 > (3^{3/2})(3^{sqrt{3x+2} - sqrt{3x}})left(dfrac{x+1}{x}right)$




    (8) So we can reject step (4).






    elementary-number-theory proof-verification prime-numbers least-common-multiple







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      2



      $begingroup$


      Let $text{lcm}(x)$ be the least common multiple of ${1,2,3,dots, x}$.



      Hanson showed that $text{lcm}(x) < 3^x$



      I'm wondering if the following argument is valid for showing that there is always a prime $p$ such that $2x < p < 3x$ for $x ge 246$.



      Here is my argument:



      (1) if prime $p$ satisfies $2x < p < 3x$, then $p | {{3x}choose{2x}}$.




      This follows from the fact that ${{3x}choose{2x}}$ is an integer and that $p$ will not divide out by $2x!$ or $x!$.




      (2) ${{3x}choose{2x}} > dfrac{6^x}{x}$ for $x ge 4$




      For $x ge 4$, ${{2x}choose{x}} ge dfrac{4^x}{x}$ since ${8choose4} = 70 > dfrac{4^4}{4} = 64$ and ${{2x}choose{x}} = 2left(dfrac{2x-1}{x}right){2(x-1)choose{x-1}} > 2left(dfrac{2x-1}{x}right)left(dfrac{4^{x-1}}{x-1}right) > dfrac{4^x}{x}$ and ${{3x}choose{2x}} > left(dfrac{3x}{2x}right)^x {{2x}choose{x}} > left(dfrac{3}{2}right)^xdfrac{4^x}{x}$




      (3) For any prime $q$ such that $frac{3x}{2} le q le 2x$, it follows that $2q > 3x$ and $q nmid {{3x}choose{2x}}$ since it will be divided out by $2x!$.



      (4) Assume that there is no prime greater than $2x$ that divides ${{3x}choose{2x}}$



      (5) It follows that ${{3x}choose{2x}} < text{lcm}(frac{3x}{2})text{lcm}(sqrt{3x})$ since:





      • From Legendre's Formula, it is well known that if $v_p(x)$ is the highest power of $p$ that divides $x$, then $v_p({{3x}choose{2x}}) = sumlimits_{i le log_p(3n)} leftlfloorfrac{3x}{p^i}rightrfloor - leftlfloorfrac{2x}{p^i}rightrfloor - leftlfloorfrac{x}{p^i}rightrfloor$ which equals $1$ or $0$ for each $i$ so that $v_p({{3x}choose{2x}}) le log_p(3x)$


      • If a prime $p > sqrt{3x}$, then $v_p({{3x}choose{2x}}) = 1$ and $p | text{lcm}(frac{3x}{2})$


      • If a prime $p le sqrt{3x}$, then $v_p({{3x}choose{2x}}) le log_p(3x) le v_p(frac{3x}{2})+1$





      (6) So that: $dfrac{6^x}{x} < 3^{3x/2}3^{sqrt{3x}}$



      (7) But this is not true for $x ge 246$ since:




      $246ln 6 - ln 246 > 435.26 > 435.24 > frac{3}{2}(246)ln 3 + sqrt{3times246}ln 3$



      and for $x ge 246$, $6 > (3^{3/2})(3^{sqrt{3x+2} - sqrt{3x}})left(dfrac{x+1}{x}right)$




      (8) So we can reject step (4).






      elementary-number-theory proof-verification prime-numbers least-common-multiple







      share|cite|improve this question









      $endgroup$




      Let $text{lcm}(x)$ be the least common multiple of ${1,2,3,dots, x}$.



      Hanson showed that $text{lcm}(x) < 3^x$



      I'm wondering if the following argument is valid for showing that there is always a prime $p$ such that $2x < p < 3x$ for $x ge 246$.



      Here is my argument:



      (1) if prime $p$ satisfies $2x < p < 3x$, then $p | {{3x}choose{2x}}$.




      This follows from the fact that ${{3x}choose{2x}}$ is an integer and that $p$ will not divide out by $2x!$ or $x!$.




      (2) ${{3x}choose{2x}} > dfrac{6^x}{x}$ for $x ge 4$




      For $x ge 4$, ${{2x}choose{x}} ge dfrac{4^x}{x}$ since ${8choose4} = 70 > dfrac{4^4}{4} = 64$ and ${{2x}choose{x}} = 2left(dfrac{2x-1}{x}right){2(x-1)choose{x-1}} > 2left(dfrac{2x-1}{x}right)left(dfrac{4^{x-1}}{x-1}right) > dfrac{4^x}{x}$ and ${{3x}choose{2x}} > left(dfrac{3x}{2x}right)^x {{2x}choose{x}} > left(dfrac{3}{2}right)^xdfrac{4^x}{x}$




      (3) For any prime $q$ such that $frac{3x}{2} le q le 2x$, it follows that $2q > 3x$ and $q nmid {{3x}choose{2x}}$ since it will be divided out by $2x!$.



      (4) Assume that there is no prime greater than $2x$ that divides ${{3x}choose{2x}}$



      (5) It follows that ${{3x}choose{2x}} < text{lcm}(frac{3x}{2})text{lcm}(sqrt{3x})$ since:





      • From Legendre's Formula, it is well known that if $v_p(x)$ is the highest power of $p$ that divides $x$, then $v_p({{3x}choose{2x}}) = sumlimits_{i le log_p(3n)} leftlfloorfrac{3x}{p^i}rightrfloor - leftlfloorfrac{2x}{p^i}rightrfloor - leftlfloorfrac{x}{p^i}rightrfloor$ which equals $1$ or $0$ for each $i$ so that $v_p({{3x}choose{2x}}) le log_p(3x)$


      • If a prime $p > sqrt{3x}$, then $v_p({{3x}choose{2x}}) = 1$ and $p | text{lcm}(frac{3x}{2})$


      • If a prime $p le sqrt{3x}$, then $v_p({{3x}choose{2x}}) le log_p(3x) le v_p(frac{3x}{2})+1$





      (6) So that: $dfrac{6^x}{x} < 3^{3x/2}3^{sqrt{3x}}$



      (7) But this is not true for $x ge 246$ since:




      $246ln 6 - ln 246 > 435.26 > 435.24 > frac{3}{2}(246)ln 3 + sqrt{3times246}ln 3$



      and for $x ge 246$, $6 > (3^{3/2})(3^{sqrt{3x+2} - sqrt{3x}})left(dfrac{x+1}{x}right)$




      (8) So we can reject step (4).






      elementary-number-theory proof-verification prime-numbers least-common-multiple




      elementary-number-theory proof-verification prime-numbers least-common-multiple






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      asked Dec 9 '18 at 8:56









      Larry FreemanLarry Freeman

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