Using Least common multiple to establish a lower bound for a ratio of primorials












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$begingroup$


Let $x#$ be the primorial for $x$.



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



It occurs to me that for $x ge 4$, it is straight forward to find a lower bound of $dfrac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}$.



I find it interesting that an upper bound can lead to a lower bound.



Is my reasoning correct?



Here's my argument:



(1) $text{lcm}left(dfrac{x^2+x}{2}right)text{lcm}(x)left(dfrac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}right) > text{lcm}(x^2+x) $





  • if a prime $p ge (x+1)$, then $p^2 > (x^2+x)$


  • if a prime $(x^2+x) ge p^a > frac{x^2+x}{2}$, then $p^{a+1} > (x^2+x)$ and $p^{a-1} le frac{x^2+x}{2}$





(2) $text{lcm}(x^2+x) ge {{x^2+x}choose{frac{x^2+x}{2}}}$




This follows from Legendre's Formula since:




  • Let $v_p(x)$ be the highest power of $p$ that divides $x$


  • $v_p({{x^2+x}choose{frac{x^2+x}{2}}}) = sumlimits_{i ge 1}leftlfloordfrac{x^2+x}{p^i}rightrfloor - 2leftlfloordfrac{x^2+x}{2p^i}rightrfloor$


  • It is well known that for each $i$, the difference is at most $1$ and that if $i > log_p(x^2+x)$, then the difference is $0$.





(3) ${{x^2+x}choose{frac{x^2+x}{2}}} > dfrac{2^{x^2+x}}{frac{x^2+x}{2}}$




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




(4) Using Hanson's result that $text{lcm}(x) < 3^x$, gives:



$$left(frac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}right) > left(frac{4}{3}right)^{(x^2+x)/2}left(frac{2}{(x^2+x)3^x}right)$$






proof-verification prime-numbers upper-lower-bounds least-common-multiple







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    0












    $begingroup$


    Let $x#$ be the primorial for $x$.



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



    It occurs to me that for $x ge 4$, it is straight forward to find a lower bound of $dfrac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}$.



    I find it interesting that an upper bound can lead to a lower bound.



    Is my reasoning correct?



    Here's my argument:



    (1) $text{lcm}left(dfrac{x^2+x}{2}right)text{lcm}(x)left(dfrac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}right) > text{lcm}(x^2+x) $





    • if a prime $p ge (x+1)$, then $p^2 > (x^2+x)$


    • if a prime $(x^2+x) ge p^a > frac{x^2+x}{2}$, then $p^{a+1} > (x^2+x)$ and $p^{a-1} le frac{x^2+x}{2}$





    (2) $text{lcm}(x^2+x) ge {{x^2+x}choose{frac{x^2+x}{2}}}$




    This follows from Legendre's Formula since:




    • Let $v_p(x)$ be the highest power of $p$ that divides $x$


    • $v_p({{x^2+x}choose{frac{x^2+x}{2}}}) = sumlimits_{i ge 1}leftlfloordfrac{x^2+x}{p^i}rightrfloor - 2leftlfloordfrac{x^2+x}{2p^i}rightrfloor$


    • It is well known that for each $i$, the difference is at most $1$ and that if $i > log_p(x^2+x)$, then the difference is $0$.





    (3) ${{x^2+x}choose{frac{x^2+x}{2}}} > dfrac{2^{x^2+x}}{frac{x^2+x}{2}}$




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




    (4) Using Hanson's result that $text{lcm}(x) < 3^x$, gives:



    $$left(frac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}right) > left(frac{4}{3}right)^{(x^2+x)/2}left(frac{2}{(x^2+x)3^x}right)$$






    proof-verification prime-numbers upper-lower-bounds least-common-multiple







    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Let $x#$ be the primorial for $x$.



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



      It occurs to me that for $x ge 4$, it is straight forward to find a lower bound of $dfrac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}$.



      I find it interesting that an upper bound can lead to a lower bound.



      Is my reasoning correct?



      Here's my argument:



      (1) $text{lcm}left(dfrac{x^2+x}{2}right)text{lcm}(x)left(dfrac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}right) > text{lcm}(x^2+x) $





      • if a prime $p ge (x+1)$, then $p^2 > (x^2+x)$


      • if a prime $(x^2+x) ge p^a > frac{x^2+x}{2}$, then $p^{a+1} > (x^2+x)$ and $p^{a-1} le frac{x^2+x}{2}$





      (2) $text{lcm}(x^2+x) ge {{x^2+x}choose{frac{x^2+x}{2}}}$




      This follows from Legendre's Formula since:




      • Let $v_p(x)$ be the highest power of $p$ that divides $x$


      • $v_p({{x^2+x}choose{frac{x^2+x}{2}}}) = sumlimits_{i ge 1}leftlfloordfrac{x^2+x}{p^i}rightrfloor - 2leftlfloordfrac{x^2+x}{2p^i}rightrfloor$


      • It is well known that for each $i$, the difference is at most $1$ and that if $i > log_p(x^2+x)$, then the difference is $0$.





      (3) ${{x^2+x}choose{frac{x^2+x}{2}}} > dfrac{2^{x^2+x}}{frac{x^2+x}{2}}$




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




      (4) Using Hanson's result that $text{lcm}(x) < 3^x$, gives:



      $$left(frac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}right) > left(frac{4}{3}right)^{(x^2+x)/2}left(frac{2}{(x^2+x)3^x}right)$$






      proof-verification prime-numbers upper-lower-bounds least-common-multiple







      share|cite|improve this question









      $endgroup$




      Let $x#$ be the primorial for $x$.



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



      It occurs to me that for $x ge 4$, it is straight forward to find a lower bound of $dfrac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}$.



      I find it interesting that an upper bound can lead to a lower bound.



      Is my reasoning correct?



      Here's my argument:



      (1) $text{lcm}left(dfrac{x^2+x}{2}right)text{lcm}(x)left(dfrac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}right) > text{lcm}(x^2+x) $





      • if a prime $p ge (x+1)$, then $p^2 > (x^2+x)$


      • if a prime $(x^2+x) ge p^a > frac{x^2+x}{2}$, then $p^{a+1} > (x^2+x)$ and $p^{a-1} le frac{x^2+x}{2}$





      (2) $text{lcm}(x^2+x) ge {{x^2+x}choose{frac{x^2+x}{2}}}$




      This follows from Legendre's Formula since:




      • Let $v_p(x)$ be the highest power of $p$ that divides $x$


      • $v_p({{x^2+x}choose{frac{x^2+x}{2}}}) = sumlimits_{i ge 1}leftlfloordfrac{x^2+x}{p^i}rightrfloor - 2leftlfloordfrac{x^2+x}{2p^i}rightrfloor$


      • It is well known that for each $i$, the difference is at most $1$ and that if $i > log_p(x^2+x)$, then the difference is $0$.





      (3) ${{x^2+x}choose{frac{x^2+x}{2}}} > dfrac{2^{x^2+x}}{frac{x^2+x}{2}}$




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




      (4) Using Hanson's result that $text{lcm}(x) < 3^x$, gives:



      $$left(frac{(x^2+x)#}{left(frac{x^2+x}{2}right)#}right) > left(frac{4}{3}right)^{(x^2+x)/2}left(frac{2}{(x^2+x)3^x}right)$$






      proof-verification prime-numbers upper-lower-bounds least-common-multiple




      proof-verification prime-numbers upper-lower-bounds least-common-multiple






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 12 '18 at 3:49









      Larry FreemanLarry Freeman

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      3,17521239






















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