Generating function of number of assignments with constraints












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Suppose I have $N$ boxes on a ring. Each box can be assigned number $0$ or $1$. Neighboring boxes can not have both $1$. so the assignment $0000$ is allowed, but the assignment $0110$, $1001$ etc are not allowed (remember the sequence is defined on a ring, so the first element and the last one are adjacent). How to find a generating function $G(q)$ that gives the number of allowed assignments $t_N$ such that
$$G(q)=sum_{N=0}^{infty} t_N q^N$$



Thanks in advance.






sequences-and-series combinatorics generating-functions fibonacci-numbers







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    Suppose I have $N$ boxes on a ring. Each box can be assigned number $0$ or $1$. Neighboring boxes can not have both $1$. so the assignment $0000$ is allowed, but the assignment $0110$, $1001$ etc are not allowed (remember the sequence is defined on a ring, so the first element and the last one are adjacent). How to find a generating function $G(q)$ that gives the number of allowed assignments $t_N$ such that
    $$G(q)=sum_{N=0}^{infty} t_N q^N$$



    Thanks in advance.






    sequences-and-series combinatorics generating-functions fibonacci-numbers







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      0












      0








      0







      Suppose I have $N$ boxes on a ring. Each box can be assigned number $0$ or $1$. Neighboring boxes can not have both $1$. so the assignment $0000$ is allowed, but the assignment $0110$, $1001$ etc are not allowed (remember the sequence is defined on a ring, so the first element and the last one are adjacent). How to find a generating function $G(q)$ that gives the number of allowed assignments $t_N$ such that
      $$G(q)=sum_{N=0}^{infty} t_N q^N$$



      Thanks in advance.






      sequences-and-series combinatorics generating-functions fibonacci-numbers







      share|cite|improve this question













      Suppose I have $N$ boxes on a ring. Each box can be assigned number $0$ or $1$. Neighboring boxes can not have both $1$. so the assignment $0000$ is allowed, but the assignment $0110$, $1001$ etc are not allowed (remember the sequence is defined on a ring, so the first element and the last one are adjacent). How to find a generating function $G(q)$ that gives the number of allowed assignments $t_N$ such that
      $$G(q)=sum_{N=0}^{infty} t_N q^N$$



      Thanks in advance.






      sequences-and-series combinatorics generating-functions fibonacci-numbers




      sequences-and-series combinatorics generating-functions fibonacci-numbers






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      asked Nov 29 '18 at 2:44









      user34104

      1478




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          These numbers have a close relationship to Fibonacci numbers. Indeed, if $f_n$ denotes the number of ${0,1}$ sequences of length $n$ which have no two consecutive $1$'s, then $f_n$ is the $n$th Fibonacci number (starting with $f_0=1,f_1=2$.



          To see the relationship between $t_n$ and $f_n$, consider a ${0,1}$ ring of length $n$ and consider the two possibilities for the first entry:




          • If the first entry is a $1$, then both the second and last entries must be $0$'s. Thus, upon deleting these three entries, we're left with a ${0,1}$ sequence of length $n-3$ which has no two consecutive $1$'s. Furthermore, this process is easily reversed, so there are precisely $f_{n-3}$ rings of this type.

          • If the first entry is a $0$, then we have no restriction on the second and last entries. Hence, removing the first entry, we're left with a ${0,1}$ sequence of length $n-1$ which has no two consecutive $1$'s. Since this process is also reversible, we have precisely $f_{n-1}$ rings of this type.


          Putting this together, we have $t_n=f_{n-1}+f_{n-3}$ for $ngeq 3$. We can also explicitly determine $t_0=1,t_1=2,t_2=3$. Now, let $F(q)=sum_n f_nq^n$ be the generating function for our Fibonacci sequence. We can compute,
          $$
          G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
          $$






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            1 Answer
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            active

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            1














            These numbers have a close relationship to Fibonacci numbers. Indeed, if $f_n$ denotes the number of ${0,1}$ sequences of length $n$ which have no two consecutive $1$'s, then $f_n$ is the $n$th Fibonacci number (starting with $f_0=1,f_1=2$.



            To see the relationship between $t_n$ and $f_n$, consider a ${0,1}$ ring of length $n$ and consider the two possibilities for the first entry:




            • If the first entry is a $1$, then both the second and last entries must be $0$'s. Thus, upon deleting these three entries, we're left with a ${0,1}$ sequence of length $n-3$ which has no two consecutive $1$'s. Furthermore, this process is easily reversed, so there are precisely $f_{n-3}$ rings of this type.

            • If the first entry is a $0$, then we have no restriction on the second and last entries. Hence, removing the first entry, we're left with a ${0,1}$ sequence of length $n-1$ which has no two consecutive $1$'s. Since this process is also reversible, we have precisely $f_{n-1}$ rings of this type.


            Putting this together, we have $t_n=f_{n-1}+f_{n-3}$ for $ngeq 3$. We can also explicitly determine $t_0=1,t_1=2,t_2=3$. Now, let $F(q)=sum_n f_nq^n$ be the generating function for our Fibonacci sequence. We can compute,
            $$
            G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
            $$






            share|cite|improve this answer


























              1














              These numbers have a close relationship to Fibonacci numbers. Indeed, if $f_n$ denotes the number of ${0,1}$ sequences of length $n$ which have no two consecutive $1$'s, then $f_n$ is the $n$th Fibonacci number (starting with $f_0=1,f_1=2$.



              To see the relationship between $t_n$ and $f_n$, consider a ${0,1}$ ring of length $n$ and consider the two possibilities for the first entry:




              • If the first entry is a $1$, then both the second and last entries must be $0$'s. Thus, upon deleting these three entries, we're left with a ${0,1}$ sequence of length $n-3$ which has no two consecutive $1$'s. Furthermore, this process is easily reversed, so there are precisely $f_{n-3}$ rings of this type.

              • If the first entry is a $0$, then we have no restriction on the second and last entries. Hence, removing the first entry, we're left with a ${0,1}$ sequence of length $n-1$ which has no two consecutive $1$'s. Since this process is also reversible, we have precisely $f_{n-1}$ rings of this type.


              Putting this together, we have $t_n=f_{n-1}+f_{n-3}$ for $ngeq 3$. We can also explicitly determine $t_0=1,t_1=2,t_2=3$. Now, let $F(q)=sum_n f_nq^n$ be the generating function for our Fibonacci sequence. We can compute,
              $$
              G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
              $$






              share|cite|improve this answer
























                1












                1








                1






                These numbers have a close relationship to Fibonacci numbers. Indeed, if $f_n$ denotes the number of ${0,1}$ sequences of length $n$ which have no two consecutive $1$'s, then $f_n$ is the $n$th Fibonacci number (starting with $f_0=1,f_1=2$.



                To see the relationship between $t_n$ and $f_n$, consider a ${0,1}$ ring of length $n$ and consider the two possibilities for the first entry:




                • If the first entry is a $1$, then both the second and last entries must be $0$'s. Thus, upon deleting these three entries, we're left with a ${0,1}$ sequence of length $n-3$ which has no two consecutive $1$'s. Furthermore, this process is easily reversed, so there are precisely $f_{n-3}$ rings of this type.

                • If the first entry is a $0$, then we have no restriction on the second and last entries. Hence, removing the first entry, we're left with a ${0,1}$ sequence of length $n-1$ which has no two consecutive $1$'s. Since this process is also reversible, we have precisely $f_{n-1}$ rings of this type.


                Putting this together, we have $t_n=f_{n-1}+f_{n-3}$ for $ngeq 3$. We can also explicitly determine $t_0=1,t_1=2,t_2=3$. Now, let $F(q)=sum_n f_nq^n$ be the generating function for our Fibonacci sequence. We can compute,
                $$
                G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
                $$






                share|cite|improve this answer












                These numbers have a close relationship to Fibonacci numbers. Indeed, if $f_n$ denotes the number of ${0,1}$ sequences of length $n$ which have no two consecutive $1$'s, then $f_n$ is the $n$th Fibonacci number (starting with $f_0=1,f_1=2$.



                To see the relationship between $t_n$ and $f_n$, consider a ${0,1}$ ring of length $n$ and consider the two possibilities for the first entry:




                • If the first entry is a $1$, then both the second and last entries must be $0$'s. Thus, upon deleting these three entries, we're left with a ${0,1}$ sequence of length $n-3$ which has no two consecutive $1$'s. Furthermore, this process is easily reversed, so there are precisely $f_{n-3}$ rings of this type.

                • If the first entry is a $0$, then we have no restriction on the second and last entries. Hence, removing the first entry, we're left with a ${0,1}$ sequence of length $n-1$ which has no two consecutive $1$'s. Since this process is also reversible, we have precisely $f_{n-1}$ rings of this type.


                Putting this together, we have $t_n=f_{n-1}+f_{n-3}$ for $ngeq 3$. We can also explicitly determine $t_0=1,t_1=2,t_2=3$. Now, let $F(q)=sum_n f_nq^n$ be the generating function for our Fibonacci sequence. We can compute,
                $$
                G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
                $$







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                answered Nov 29 '18 at 21:47









                munchhausen

                79416




                79416






























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