How many pairs $(x,y)$ such that $x+y=n$ have an $x$ or $y$ divisible by 3 but $x$ and $y$ are not equal to...












2












$begingroup$


Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+y=n$ where $mathbb{O}$ is set of odd integers > 1.



Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.



Examples: $S_{10} ={(3,7),(5,5),(7,3)}$ and $f(10) = 3$.



$S_{14} = {(3,11), (5,9), (7,7), (9,5),(11,3)}$ and $f(14) = 5$



$S_{32} = {(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)}$



and $f(32) = 14$



It turns out that $f(n) = ((n/2) - 2)$.



Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x neq 3$ and $yneq 3$.



Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).



There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.



What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?



What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?



My guess is that as $n to infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.






number-theory goldbachs-conjecture







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+y=n$ where $mathbb{O}$ is set of odd integers > 1.



    Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.



    Examples: $S_{10} ={(3,7),(5,5),(7,3)}$ and $f(10) = 3$.



    $S_{14} = {(3,11), (5,9), (7,7), (9,5),(11,3)}$ and $f(14) = 5$



    $S_{32} = {(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)}$



    and $f(32) = 14$



    It turns out that $f(n) = ((n/2) - 2)$.



    Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x neq 3$ and $yneq 3$.



    Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
    They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).



    There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.



    What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?



    What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?



    My guess is that as $n to infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.






    number-theory goldbachs-conjecture







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+y=n$ where $mathbb{O}$ is set of odd integers > 1.



      Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.



      Examples: $S_{10} ={(3,7),(5,5),(7,3)}$ and $f(10) = 3$.



      $S_{14} = {(3,11), (5,9), (7,7), (9,5),(11,3)}$ and $f(14) = 5$



      $S_{32} = {(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)}$



      and $f(32) = 14$



      It turns out that $f(n) = ((n/2) - 2)$.



      Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x neq 3$ and $yneq 3$.



      Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
      They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).



      There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.



      What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?



      What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?



      My guess is that as $n to infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.






      number-theory goldbachs-conjecture







      share|cite|improve this question









      $endgroup$




      Let the set $S_{n}$ = {$(x,y):x,y in mathbb{O}$} such that $x+y=n$ where $mathbb{O}$ is set of odd integers > 1.



      Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.



      Examples: $S_{10} ={(3,7),(5,5),(7,3)}$ and $f(10) = 3$.



      $S_{14} = {(3,11), (5,9), (7,7), (9,5),(11,3)}$ and $f(14) = 5$



      $S_{32} = {(3,29), (5,27), (7,25), (9,23), (11,21), (13,19), (15,17), (17,15), (19,13), (21,11), (23,9), (25,7), (27,5), (29,3)}$



      and $f(32) = 14$



      It turns out that $f(n) = ((n/2) - 2)$.



      Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 3, but $x neq 3$ and $yneq 3$.



      Example: $g(32) = 8$ because there are 8 pairs where $x$ or $y$ is divisible by 3 but not equal to 3.
      They are (5,27), (9,23), (11,21), (15,17), (17,15), (21,11), (23,9) and (27,5).



      There are two cases: Case 1, $n$ is divisible by 3 and case 2, $n$ is not divisible by 3. Let us only consider case where $n$ is not divisible by 3, since if $n$ is divisible by 3, any pair where $x$ is divisible by 3, $y$ will also be divisible by 3.



      What is the formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by 3?



      What is the formula for $g(n)$ in terms of $f(n)$ for limit $n toinfty$?



      My guess is that as $n to infty$ for $n$ not divisible by 3, the value of $g(n)$ approaches $(2/3)f(n)$.






      number-theory goldbachs-conjecture




      number-theory goldbachs-conjecture






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 13 '18 at 16:29









      temp wattstemp watts

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          $begingroup$

          The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.



          So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:



          $$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$



          The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:



          $$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$



          Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:



          $$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$



          $$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$






          share|cite|improve this answer









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            $begingroup$

            The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.



            So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:



            $$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$



            The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:



            $$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$



            Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:



            $$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$



            $$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$






            share|cite|improve this answer









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              0












              $begingroup$

              The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.



              So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:



              $$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$



              The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:



              $$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$



              Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:



              $$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$



              $$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.



                So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:



                $$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$



                The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:



                $$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$



                Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:



                $$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$



                $$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$






                share|cite|improve this answer









                $endgroup$



                The sum of $2$ multiples of $k$ is also a multiple of $k$. Therefore, if $n$ is not divisible by $3$, we know that the $leftlfloor frac{n}{6}rightrfloor -1$ values of $p$ where $p$ is divisible by $3$ will not overlap any of the $leftlfloor frac{n}{6}rightrfloor -1$ values of $q$ where $q$ is divisible by $3$.



                So the formula for $g(n)$ in terms of $n$ for values of $n$ not divisible by $3$ is:



                $$g(n)=2leftlfloor frac{n}{6}rightrfloor -2$$



                The formula for $g(n)$ in terms of $f(n)$ for values of $n$ not divisible by $3$ is:



                $$g(n)=2leftlfloor frac{fleft(nright)+2}{3}rightrfloor -2$$



                Your guess about the limit for $frac{g(n)}{fleft(nright)}$ is also right:



                $$frac{g(n)}{fleft(nright)}=frac{2leftlfloor frac{n}{6}rightrfloor -2}{frac{n}{2}-2}=frac{frac{n}{3}-2}{frac{n}{2}-2}=frac{2n-12}{3n-12}$$



                $$lim_{nrightarrowinfty}frac{2n-12}{3n-12}=frac{2infty-12}{3infty-12}=frac{2infty}{3infty}=frac{2}{3}left(frac{infty}{infty}right)=frac{2}{3}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 23 '18 at 1:57









                François HuppéFrançois Huppé

                357111




                357111






























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