Least Common Multiple and the product of a sequence of consecutive integers

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Let $x>0, n>0$ be integers.



Let LCM$(x+1, x+2, dots, x+n)$ be the least common multiple of $x+1, x+2, dots, x+n$.



It seems to me that:



$$frac(x+1)(x+2)timesdotstimes(x+n)textLCM(x+1, x+2, dots, x+n) le (n-1)!$$



Here is my thinking:



  • Let $p$ be any prime that divides $(x+1)timesdotstimes(x+n)$.


  • Let $x+i$ be the integer in $x+1,dots, x+n$ that is divisible by the highest power of $p$.


  • All integers in the sequence $x+1, dots, x+n$ that are divisible by $p$ will have the form $x+i+ap$ or $x+i-ap$ where $a>0$ is an integer where $x+1 le x+i - ap le x+n$ or $x+1 le x+i+ap le x+n$.


  • This means that the highest $a$ will be $lfloorfracn-1prfloor$


  • So, it follows that the maximum power of $p$ that divides $dfrac(x+1)timesdotstimes(x+n)textLCM(x+1,dots,x+n)$ is the same as the maximum power of $p$ that divides $(n-1)!$






elementary-number-theory inequality prime-numbers least-common-multiple







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    Let $x>0, n>0$ be integers.



    Let LCM$(x+1, x+2, dots, x+n)$ be the least common multiple of $x+1, x+2, dots, x+n$.



    It seems to me that:



    $$frac(x+1)(x+2)timesdotstimes(x+n)textLCM(x+1, x+2, dots, x+n) le (n-1)!$$



    Here is my thinking:



    • Let $p$ be any prime that divides $(x+1)timesdotstimes(x+n)$.


    • Let $x+i$ be the integer in $x+1,dots, x+n$ that is divisible by the highest power of $p$.


    • All integers in the sequence $x+1, dots, x+n$ that are divisible by $p$ will have the form $x+i+ap$ or $x+i-ap$ where $a>0$ is an integer where $x+1 le x+i - ap le x+n$ or $x+1 le x+i+ap le x+n$.


    • This means that the highest $a$ will be $lfloorfracn-1prfloor$


    • So, it follows that the maximum power of $p$ that divides $dfrac(x+1)timesdotstimes(x+n)textLCM(x+1,dots,x+n)$ is the same as the maximum power of $p$ that divides $(n-1)!$






    elementary-number-theory inequality prime-numbers least-common-multiple







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      Let $x>0, n>0$ be integers.



      Let LCM$(x+1, x+2, dots, x+n)$ be the least common multiple of $x+1, x+2, dots, x+n$.



      It seems to me that:



      $$frac(x+1)(x+2)timesdotstimes(x+n)textLCM(x+1, x+2, dots, x+n) le (n-1)!$$



      Here is my thinking:



      • Let $p$ be any prime that divides $(x+1)timesdotstimes(x+n)$.


      • Let $x+i$ be the integer in $x+1,dots, x+n$ that is divisible by the highest power of $p$.


      • All integers in the sequence $x+1, dots, x+n$ that are divisible by $p$ will have the form $x+i+ap$ or $x+i-ap$ where $a>0$ is an integer where $x+1 le x+i - ap le x+n$ or $x+1 le x+i+ap le x+n$.


      • This means that the highest $a$ will be $lfloorfracn-1prfloor$


      • So, it follows that the maximum power of $p$ that divides $dfrac(x+1)timesdotstimes(x+n)textLCM(x+1,dots,x+n)$ is the same as the maximum power of $p$ that divides $(n-1)!$






      elementary-number-theory inequality prime-numbers least-common-multiple







      share|cite|improve this question













      Let $x>0, n>0$ be integers.



      Let LCM$(x+1, x+2, dots, x+n)$ be the least common multiple of $x+1, x+2, dots, x+n$.



      It seems to me that:



      $$frac(x+1)(x+2)timesdotstimes(x+n)textLCM(x+1, x+2, dots, x+n) le (n-1)!$$



      Here is my thinking:



      • Let $p$ be any prime that divides $(x+1)timesdotstimes(x+n)$.


      • Let $x+i$ be the integer in $x+1,dots, x+n$ that is divisible by the highest power of $p$.


      • All integers in the sequence $x+1, dots, x+n$ that are divisible by $p$ will have the form $x+i+ap$ or $x+i-ap$ where $a>0$ is an integer where $x+1 le x+i - ap le x+n$ or $x+1 le x+i+ap le x+n$.


      • This means that the highest $a$ will be $lfloorfracn-1prfloor$


      • So, it follows that the maximum power of $p$ that divides $dfrac(x+1)timesdotstimes(x+n)textLCM(x+1,dots,x+n)$ is the same as the maximum power of $p$ that divides $(n-1)!$






      elementary-number-theory inequality prime-numbers least-common-multiple




      elementary-number-theory inequality prime-numbers least-common-multiple






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      asked Sep 6 at 5:23









      Larry Freeman

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