Solve the recurrence relation $n(n-1)a_n - (n-2)^2a_{n-2}= 0$, $a_0=0$, $a_1=1$












0















Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.




I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.






sequences-and-series combinatorics discrete-mathematics recurrence-relations







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  • Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
    – José Carlos Santos
    Nov 24 at 21:05










  • @JoséCarlosSantos Yes, that is what I was given.
    – cosmicbrownie
    Nov 24 at 21:16
















0















Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.




I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.






sequences-and-series combinatorics discrete-mathematics recurrence-relations







share|cite|improve this question
























  • Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
    – José Carlos Santos
    Nov 24 at 21:05










  • @JoséCarlosSantos Yes, that is what I was given.
    – cosmicbrownie
    Nov 24 at 21:16














0












0








0








Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.




I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.






sequences-and-series combinatorics discrete-mathematics recurrence-relations







share|cite|improve this question
















Solve the recurrence relation $$n(n-1)a_n - (n-2)^2a_{n-2}= 0,$$ where $a_0=0$, $a_1=1$.




I think I might need to use generating functions, but I'm still not sure how to get started with this problem. Typing this into software came back as no solution for some reason.






sequences-and-series combinatorics discrete-mathematics recurrence-relations




sequences-and-series combinatorics discrete-mathematics recurrence-relations






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share|cite|improve this question













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edited Nov 24 at 21:17









greedoid

37.2k114794




37.2k114794










asked Nov 24 at 21:00









cosmicbrownie

1016




1016












  • Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
    – José Carlos Santos
    Nov 24 at 21:05










  • @JoséCarlosSantos Yes, that is what I was given.
    – cosmicbrownie
    Nov 24 at 21:16


















  • Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
    – José Carlos Santos
    Nov 24 at 21:05










  • @JoséCarlosSantos Yes, that is what I was given.
    – cosmicbrownie
    Nov 24 at 21:16
















Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
– José Carlos Santos
Nov 24 at 21:05




Are you sure about the definition of the $a_n$'s? With that definition, it is trivial that $a_n=0$ when $n$ is even.
– José Carlos Santos
Nov 24 at 21:05












@JoséCarlosSantos Yes, that is what I was given.
– cosmicbrownie
Nov 24 at 21:16




@JoséCarlosSantos Yes, that is what I was given.
– cosmicbrownie
Nov 24 at 21:16










2 Answers
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As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
$$a_n = {((n-2)!!)^2over n!}$$



You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).






share|cite|improve this answer































    0














    $a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$



    Clearly, all the even-index terms of the recurrence are $0$.
    For the odd-index terms, you could do this:



    $a_3=frac{1^2a_1}{2cdot3}\
    a_5=frac{3^2a_3}{4cdot5}\
    a_7=frac{5^2a_5}{6cdot7}\
    cdot\
    cdot\
    cdot\
    a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
    $



    Multiplying all the terms
    $thinspacethinspace
    implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
    implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
    implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
    implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
    $






    share|cite|improve this answer























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      2














      As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
      $$a_n = {((n-2)!!)^2over n!}$$



      You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).






      share|cite|improve this answer




























        2














        As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
        $$a_n = {((n-2)!!)^2over n!}$$



        You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).






        share|cite|improve this answer


























          2












          2








          2






          As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
          $$a_n = {((n-2)!!)^2over n!}$$



          You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).






          share|cite|improve this answer














          As José said for $n$ even $a_n=0$. Now for $n$ odd it is easy to show (say with inudction) that $$a_n = {1cdot 3^2cdot 5^2cdot...cdot (n-2)^2over n!}$$ or
          $$a_n = {((n-2)!!)^2over n!}$$



          You may find this formula by calculating the values of $a_n$ for small $n$ and then prove it (as I said by induction).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 21:19

























          answered Nov 24 at 21:12









          greedoid

          37.2k114794




          37.2k114794























              0














              $a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$



              Clearly, all the even-index terms of the recurrence are $0$.
              For the odd-index terms, you could do this:



              $a_3=frac{1^2a_1}{2cdot3}\
              a_5=frac{3^2a_3}{4cdot5}\
              a_7=frac{5^2a_5}{6cdot7}\
              cdot\
              cdot\
              cdot\
              a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
              $



              Multiplying all the terms
              $thinspacethinspace
              implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
              implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
              implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
              implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
              $






              share|cite|improve this answer




























                0














                $a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$



                Clearly, all the even-index terms of the recurrence are $0$.
                For the odd-index terms, you could do this:



                $a_3=frac{1^2a_1}{2cdot3}\
                a_5=frac{3^2a_3}{4cdot5}\
                a_7=frac{5^2a_5}{6cdot7}\
                cdot\
                cdot\
                cdot\
                a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
                $



                Multiplying all the terms
                $thinspacethinspace
                implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
                implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
                implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
                implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
                $






                share|cite|improve this answer


























                  0












                  0








                  0






                  $a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$



                  Clearly, all the even-index terms of the recurrence are $0$.
                  For the odd-index terms, you could do this:



                  $a_3=frac{1^2a_1}{2cdot3}\
                  a_5=frac{3^2a_3}{4cdot5}\
                  a_7=frac{5^2a_5}{6cdot7}\
                  cdot\
                  cdot\
                  cdot\
                  a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
                  $



                  Multiplying all the terms
                  $thinspacethinspace
                  implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
                  implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
                  implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
                  implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
                  $






                  share|cite|improve this answer














                  $a_n=frac{(n-2)^2a_{n-2}}{n(n-1)}, ngeq2; a_0=0, a_1=1$



                  Clearly, all the even-index terms of the recurrence are $0$.
                  For the odd-index terms, you could do this:



                  $a_3=frac{1^2a_1}{2cdot3}\
                  a_5=frac{3^2a_3}{4cdot5}\
                  a_7=frac{5^2a_5}{6cdot7}\
                  cdot\
                  cdot\
                  cdot\
                  a_{2n+1}=frac{(2n-1)^2a_{2n-1}}{(2n)cdot(2n+1)}, ngeq1\
                  $



                  Multiplying all the terms
                  $thinspacethinspace
                  implies a_3cdot a_5cdot a_7 cdot cdot cdot a_{2n+1}=frac{1^2cdot3^2cdot5^2cdotcdotcdot(2n-1)^2}{(2n+1)!}a_1cdot a_3cdot a_5cdotcdotcdot a_{2n-1}\
                  implies a_{2n+1}=frac{(1cdot3cdot5cdotcdotcdot(2n-1))^2}{(2n+1)!}a_1\
                  implies a_{2n+1}=frac{(2n-1)!^2}{2^{2n-2} (n-1)!^2 (2n+1)!}\
                  implies a_{2n+1}=frac{(2n)!}{2^{2n} (2n+1) n!^2}; ngeq1
                  $







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 24 at 21:44

























                  answered Nov 24 at 21:31









                  Shubham Johri

                  3,273615




                  3,273615






























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