Formula for $a_n$ where $a_n$ = n*($a_{n-1}$+$a_{n-2}$)












4












$begingroup$


Formula for $a_n$ if $a_0=1$ and $a_1=2$ and n greater than equal to 2



$a_n$ = $n times (a_{n-1}+a_{n-2}$).



Attempt: $ a_2 =6, a_3=24, a_4=120, a_5=720 , a_6 = 5040 $ It so looks like $a_n= (n+1)!$



I tried to open $n times (n-1 times (a_{n-2}+a_{n-3}) + a_{n-2}$).
Unable to get formula in factorial terms.






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: Have you tried using induction?
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:15










  • $begingroup$
    Using induction seems like cheating. I was thinking more in terms of generating functions, once unable to do directly by splitting.
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:34






  • 1




    $begingroup$
    Another way to look at it is that all you know about the sequence $a_n$ is an inductive definition, so you have to use induction to prove anything about it. The use of induction might be implicit, but it's there somewhere. (If you wrote down all the formal details of a proof using generating functions, you'd have to use induction; if your proof uses $"ldots"$ anywhere, a formal proof would use induction; etc.)
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:39










  • $begingroup$
    Wow! thanks, didn't know that. Always thought generating functions was the superior way to do recurrence relations.
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:46






  • 1




    $begingroup$
    Generating functions are fine, and they can be a great way to discover and organize a proof. But fundamentally they aren't an alternative to induction; they're really a way to handle the bookkeeping of an induction automatically (and also to connect things with concepts of analysis, which can be useful). So I certainly don't mean to discourage the use of generating functions!
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:48
















4












$begingroup$


Formula for $a_n$ if $a_0=1$ and $a_1=2$ and n greater than equal to 2



$a_n$ = $n times (a_{n-1}+a_{n-2}$).



Attempt: $ a_2 =6, a_3=24, a_4=120, a_5=720 , a_6 = 5040 $ It so looks like $a_n= (n+1)!$



I tried to open $n times (n-1 times (a_{n-2}+a_{n-3}) + a_{n-2}$).
Unable to get formula in factorial terms.






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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: Have you tried using induction?
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:15










  • $begingroup$
    Using induction seems like cheating. I was thinking more in terms of generating functions, once unable to do directly by splitting.
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:34






  • 1




    $begingroup$
    Another way to look at it is that all you know about the sequence $a_n$ is an inductive definition, so you have to use induction to prove anything about it. The use of induction might be implicit, but it's there somewhere. (If you wrote down all the formal details of a proof using generating functions, you'd have to use induction; if your proof uses $"ldots"$ anywhere, a formal proof would use induction; etc.)
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:39










  • $begingroup$
    Wow! thanks, didn't know that. Always thought generating functions was the superior way to do recurrence relations.
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:46






  • 1




    $begingroup$
    Generating functions are fine, and they can be a great way to discover and organize a proof. But fundamentally they aren't an alternative to induction; they're really a way to handle the bookkeeping of an induction automatically (and also to connect things with concepts of analysis, which can be useful). So I certainly don't mean to discourage the use of generating functions!
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:48














4












4








4


2



$begingroup$


Formula for $a_n$ if $a_0=1$ and $a_1=2$ and n greater than equal to 2



$a_n$ = $n times (a_{n-1}+a_{n-2}$).



Attempt: $ a_2 =6, a_3=24, a_4=120, a_5=720 , a_6 = 5040 $ It so looks like $a_n= (n+1)!$



I tried to open $n times (n-1 times (a_{n-2}+a_{n-3}) + a_{n-2}$).
Unable to get formula in factorial terms.






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







share|cite|improve this question









$endgroup$




Formula for $a_n$ if $a_0=1$ and $a_1=2$ and n greater than equal to 2



$a_n$ = $n times (a_{n-1}+a_{n-2}$).



Attempt: $ a_2 =6, a_3=24, a_4=120, a_5=720 , a_6 = 5040 $ It so looks like $a_n= (n+1)!$



I tried to open $n times (n-1 times (a_{n-2}+a_{n-3}) + a_{n-2}$).
Unable to get formula in factorial terms.






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




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 19 '16 at 4:09









AmritaAmrita

435314




435314












  • $begingroup$
    Hint: Have you tried using induction?
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:15










  • $begingroup$
    Using induction seems like cheating. I was thinking more in terms of generating functions, once unable to do directly by splitting.
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:34






  • 1




    $begingroup$
    Another way to look at it is that all you know about the sequence $a_n$ is an inductive definition, so you have to use induction to prove anything about it. The use of induction might be implicit, but it's there somewhere. (If you wrote down all the formal details of a proof using generating functions, you'd have to use induction; if your proof uses $"ldots"$ anywhere, a formal proof would use induction; etc.)
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:39










  • $begingroup$
    Wow! thanks, didn't know that. Always thought generating functions was the superior way to do recurrence relations.
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:46






  • 1




    $begingroup$
    Generating functions are fine, and they can be a great way to discover and organize a proof. But fundamentally they aren't an alternative to induction; they're really a way to handle the bookkeeping of an induction automatically (and also to connect things with concepts of analysis, which can be useful). So I certainly don't mean to discourage the use of generating functions!
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:48


















  • $begingroup$
    Hint: Have you tried using induction?
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:15










  • $begingroup$
    Using induction seems like cheating. I was thinking more in terms of generating functions, once unable to do directly by splitting.
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:34






  • 1




    $begingroup$
    Another way to look at it is that all you know about the sequence $a_n$ is an inductive definition, so you have to use induction to prove anything about it. The use of induction might be implicit, but it's there somewhere. (If you wrote down all the formal details of a proof using generating functions, you'd have to use induction; if your proof uses $"ldots"$ anywhere, a formal proof would use induction; etc.)
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:39










  • $begingroup$
    Wow! thanks, didn't know that. Always thought generating functions was the superior way to do recurrence relations.
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:46






  • 1




    $begingroup$
    Generating functions are fine, and they can be a great way to discover and organize a proof. But fundamentally they aren't an alternative to induction; they're really a way to handle the bookkeeping of an induction automatically (and also to connect things with concepts of analysis, which can be useful). So I certainly don't mean to discourage the use of generating functions!
    $endgroup$
    – Mitchell Spector
    Sep 19 '16 at 4:48
















$begingroup$
Hint: Have you tried using induction?
$endgroup$
– Mitchell Spector
Sep 19 '16 at 4:15




$begingroup$
Hint: Have you tried using induction?
$endgroup$
– Mitchell Spector
Sep 19 '16 at 4:15












$begingroup$
Using induction seems like cheating. I was thinking more in terms of generating functions, once unable to do directly by splitting.
$endgroup$
– Amrita
Sep 19 '16 at 4:34




$begingroup$
Using induction seems like cheating. I was thinking more in terms of generating functions, once unable to do directly by splitting.
$endgroup$
– Amrita
Sep 19 '16 at 4:34




1




1




$begingroup$
Another way to look at it is that all you know about the sequence $a_n$ is an inductive definition, so you have to use induction to prove anything about it. The use of induction might be implicit, but it's there somewhere. (If you wrote down all the formal details of a proof using generating functions, you'd have to use induction; if your proof uses $"ldots"$ anywhere, a formal proof would use induction; etc.)
$endgroup$
– Mitchell Spector
Sep 19 '16 at 4:39




$begingroup$
Another way to look at it is that all you know about the sequence $a_n$ is an inductive definition, so you have to use induction to prove anything about it. The use of induction might be implicit, but it's there somewhere. (If you wrote down all the formal details of a proof using generating functions, you'd have to use induction; if your proof uses $"ldots"$ anywhere, a formal proof would use induction; etc.)
$endgroup$
– Mitchell Spector
Sep 19 '16 at 4:39












$begingroup$
Wow! thanks, didn't know that. Always thought generating functions was the superior way to do recurrence relations.
$endgroup$
– Amrita
Sep 19 '16 at 4:46




$begingroup$
Wow! thanks, didn't know that. Always thought generating functions was the superior way to do recurrence relations.
$endgroup$
– Amrita
Sep 19 '16 at 4:46




1




1




$begingroup$
Generating functions are fine, and they can be a great way to discover and organize a proof. But fundamentally they aren't an alternative to induction; they're really a way to handle the bookkeeping of an induction automatically (and also to connect things with concepts of analysis, which can be useful). So I certainly don't mean to discourage the use of generating functions!
$endgroup$
– Mitchell Spector
Sep 19 '16 at 4:48




$begingroup$
Generating functions are fine, and they can be a great way to discover and organize a proof. But fundamentally they aren't an alternative to induction; they're really a way to handle the bookkeeping of an induction automatically (and also to connect things with concepts of analysis, which can be useful). So I certainly don't mean to discourage the use of generating functions!
$endgroup$
– Mitchell Spector
Sep 19 '16 at 4:48










2 Answers
2






active

oldest

votes


















3












$begingroup$

Once you've noticed the pattern, it can be proved using induction on $n$:



For the base case, $a_0=1!$, $a_1=2!$.



For the induction step, if $n>1$ and $a_k=(k+1)!$ for $0leq kleq n-1$, then
$$ a_n=n(a_{n-1}+a_{n-2})=n(n!+(n-1)!)=n(n-1)!(n+1)=(n+1)!$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was thinking in terms of generating functions. For me multiplying by $x^n$ both sides to get g.f seems to not work. Any way I can do this using generating functions. I am trying to learn g.f
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:36












  • $begingroup$
    Maybe try an exponential generating function instead? I think induction is the simplest approach.
    $endgroup$
    – carmichael561
    Sep 19 '16 at 4:42



















1












$begingroup$

There is a way to solve this using generating functions. The use of ordinary generating functions is more complicated due to convergence failure, but exponential generating functions work in this case.



Suppose $,a_n,$ is a sequence of numbers. Then $,y =E[a_n] := sum_{n=0}^infty a_n x^n/n!,$ is its exponential generating function. The two properties we need are $, E[na_n] = xy',$ and $,E[a_{n+1}] = y'.,$ The recursion is $,a_n = n times (a_{n-1}+a_{n-2}).,$



We modify the recursion to be $,a_{n+2} = (n+2)(a_{n+1} + a_n).,$ Now $E[a_{n+2}] = y'',,$
$,E[a_{n+1}+a_n] = (y+y').,$ The recursion becomes
$$ y'' = 2(y+y') + x(y+y')' = 2y + (x+2)y' + xy''. $$
Solving this differential equation with the initial conditions $,y(0) = 1,, y'(0) = 2,$ gives the solution $, y(x) = 1/(1-x)^2.,$ Now $, y = 1 + 2x + 3x^2 + cdots = sum_{n=0}^infty (n+1)x^n $ and thus $,y = E[(n+1)n!],$ or simply $,a_n = (n+1)!.$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Once you've noticed the pattern, it can be proved using induction on $n$:



    For the base case, $a_0=1!$, $a_1=2!$.



    For the induction step, if $n>1$ and $a_k=(k+1)!$ for $0leq kleq n-1$, then
    $$ a_n=n(a_{n-1}+a_{n-2})=n(n!+(n-1)!)=n(n-1)!(n+1)=(n+1)!$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was thinking in terms of generating functions. For me multiplying by $x^n$ both sides to get g.f seems to not work. Any way I can do this using generating functions. I am trying to learn g.f
      $endgroup$
      – Amrita
      Sep 19 '16 at 4:36












    • $begingroup$
      Maybe try an exponential generating function instead? I think induction is the simplest approach.
      $endgroup$
      – carmichael561
      Sep 19 '16 at 4:42
















    3












    $begingroup$

    Once you've noticed the pattern, it can be proved using induction on $n$:



    For the base case, $a_0=1!$, $a_1=2!$.



    For the induction step, if $n>1$ and $a_k=(k+1)!$ for $0leq kleq n-1$, then
    $$ a_n=n(a_{n-1}+a_{n-2})=n(n!+(n-1)!)=n(n-1)!(n+1)=(n+1)!$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I was thinking in terms of generating functions. For me multiplying by $x^n$ both sides to get g.f seems to not work. Any way I can do this using generating functions. I am trying to learn g.f
      $endgroup$
      – Amrita
      Sep 19 '16 at 4:36












    • $begingroup$
      Maybe try an exponential generating function instead? I think induction is the simplest approach.
      $endgroup$
      – carmichael561
      Sep 19 '16 at 4:42














    3












    3








    3





    $begingroup$

    Once you've noticed the pattern, it can be proved using induction on $n$:



    For the base case, $a_0=1!$, $a_1=2!$.



    For the induction step, if $n>1$ and $a_k=(k+1)!$ for $0leq kleq n-1$, then
    $$ a_n=n(a_{n-1}+a_{n-2})=n(n!+(n-1)!)=n(n-1)!(n+1)=(n+1)!$$






    share|cite|improve this answer









    $endgroup$



    Once you've noticed the pattern, it can be proved using induction on $n$:



    For the base case, $a_0=1!$, $a_1=2!$.



    For the induction step, if $n>1$ and $a_k=(k+1)!$ for $0leq kleq n-1$, then
    $$ a_n=n(a_{n-1}+a_{n-2})=n(n!+(n-1)!)=n(n-1)!(n+1)=(n+1)!$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 19 '16 at 4:16









    carmichael561carmichael561

    47.1k54583




    47.1k54583












    • $begingroup$
      I was thinking in terms of generating functions. For me multiplying by $x^n$ both sides to get g.f seems to not work. Any way I can do this using generating functions. I am trying to learn g.f
      $endgroup$
      – Amrita
      Sep 19 '16 at 4:36












    • $begingroup$
      Maybe try an exponential generating function instead? I think induction is the simplest approach.
      $endgroup$
      – carmichael561
      Sep 19 '16 at 4:42


















    • $begingroup$
      I was thinking in terms of generating functions. For me multiplying by $x^n$ both sides to get g.f seems to not work. Any way I can do this using generating functions. I am trying to learn g.f
      $endgroup$
      – Amrita
      Sep 19 '16 at 4:36












    • $begingroup$
      Maybe try an exponential generating function instead? I think induction is the simplest approach.
      $endgroup$
      – carmichael561
      Sep 19 '16 at 4:42
















    $begingroup$
    I was thinking in terms of generating functions. For me multiplying by $x^n$ both sides to get g.f seems to not work. Any way I can do this using generating functions. I am trying to learn g.f
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:36






    $begingroup$
    I was thinking in terms of generating functions. For me multiplying by $x^n$ both sides to get g.f seems to not work. Any way I can do this using generating functions. I am trying to learn g.f
    $endgroup$
    – Amrita
    Sep 19 '16 at 4:36














    $begingroup$
    Maybe try an exponential generating function instead? I think induction is the simplest approach.
    $endgroup$
    – carmichael561
    Sep 19 '16 at 4:42




    $begingroup$
    Maybe try an exponential generating function instead? I think induction is the simplest approach.
    $endgroup$
    – carmichael561
    Sep 19 '16 at 4:42











    1












    $begingroup$

    There is a way to solve this using generating functions. The use of ordinary generating functions is more complicated due to convergence failure, but exponential generating functions work in this case.



    Suppose $,a_n,$ is a sequence of numbers. Then $,y =E[a_n] := sum_{n=0}^infty a_n x^n/n!,$ is its exponential generating function. The two properties we need are $, E[na_n] = xy',$ and $,E[a_{n+1}] = y'.,$ The recursion is $,a_n = n times (a_{n-1}+a_{n-2}).,$



    We modify the recursion to be $,a_{n+2} = (n+2)(a_{n+1} + a_n).,$ Now $E[a_{n+2}] = y'',,$
    $,E[a_{n+1}+a_n] = (y+y').,$ The recursion becomes
    $$ y'' = 2(y+y') + x(y+y')' = 2y + (x+2)y' + xy''. $$
    Solving this differential equation with the initial conditions $,y(0) = 1,, y'(0) = 2,$ gives the solution $, y(x) = 1/(1-x)^2.,$ Now $, y = 1 + 2x + 3x^2 + cdots = sum_{n=0}^infty (n+1)x^n $ and thus $,y = E[(n+1)n!],$ or simply $,a_n = (n+1)!.$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      There is a way to solve this using generating functions. The use of ordinary generating functions is more complicated due to convergence failure, but exponential generating functions work in this case.



      Suppose $,a_n,$ is a sequence of numbers. Then $,y =E[a_n] := sum_{n=0}^infty a_n x^n/n!,$ is its exponential generating function. The two properties we need are $, E[na_n] = xy',$ and $,E[a_{n+1}] = y'.,$ The recursion is $,a_n = n times (a_{n-1}+a_{n-2}).,$



      We modify the recursion to be $,a_{n+2} = (n+2)(a_{n+1} + a_n).,$ Now $E[a_{n+2}] = y'',,$
      $,E[a_{n+1}+a_n] = (y+y').,$ The recursion becomes
      $$ y'' = 2(y+y') + x(y+y')' = 2y + (x+2)y' + xy''. $$
      Solving this differential equation with the initial conditions $,y(0) = 1,, y'(0) = 2,$ gives the solution $, y(x) = 1/(1-x)^2.,$ Now $, y = 1 + 2x + 3x^2 + cdots = sum_{n=0}^infty (n+1)x^n $ and thus $,y = E[(n+1)n!],$ or simply $,a_n = (n+1)!.$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        There is a way to solve this using generating functions. The use of ordinary generating functions is more complicated due to convergence failure, but exponential generating functions work in this case.



        Suppose $,a_n,$ is a sequence of numbers. Then $,y =E[a_n] := sum_{n=0}^infty a_n x^n/n!,$ is its exponential generating function. The two properties we need are $, E[na_n] = xy',$ and $,E[a_{n+1}] = y'.,$ The recursion is $,a_n = n times (a_{n-1}+a_{n-2}).,$



        We modify the recursion to be $,a_{n+2} = (n+2)(a_{n+1} + a_n).,$ Now $E[a_{n+2}] = y'',,$
        $,E[a_{n+1}+a_n] = (y+y').,$ The recursion becomes
        $$ y'' = 2(y+y') + x(y+y')' = 2y + (x+2)y' + xy''. $$
        Solving this differential equation with the initial conditions $,y(0) = 1,, y'(0) = 2,$ gives the solution $, y(x) = 1/(1-x)^2.,$ Now $, y = 1 + 2x + 3x^2 + cdots = sum_{n=0}^infty (n+1)x^n $ and thus $,y = E[(n+1)n!],$ or simply $,a_n = (n+1)!.$






        share|cite|improve this answer









        $endgroup$



        There is a way to solve this using generating functions. The use of ordinary generating functions is more complicated due to convergence failure, but exponential generating functions work in this case.



        Suppose $,a_n,$ is a sequence of numbers. Then $,y =E[a_n] := sum_{n=0}^infty a_n x^n/n!,$ is its exponential generating function. The two properties we need are $, E[na_n] = xy',$ and $,E[a_{n+1}] = y'.,$ The recursion is $,a_n = n times (a_{n-1}+a_{n-2}).,$



        We modify the recursion to be $,a_{n+2} = (n+2)(a_{n+1} + a_n).,$ Now $E[a_{n+2}] = y'',,$
        $,E[a_{n+1}+a_n] = (y+y').,$ The recursion becomes
        $$ y'' = 2(y+y') + x(y+y')' = 2y + (x+2)y' + xy''. $$
        Solving this differential equation with the initial conditions $,y(0) = 1,, y'(0) = 2,$ gives the solution $, y(x) = 1/(1-x)^2.,$ Now $, y = 1 + 2x + 3x^2 + cdots = sum_{n=0}^infty (n+1)x^n $ and thus $,y = E[(n+1)n!],$ or simply $,a_n = (n+1)!.$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 23 '18 at 16:21









        SomosSomos

        14.5k11336




        14.5k11336






























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