Generating function of number of assignments with constraints
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
asked Nov 29 '18 at 2:44
user34104
1478
add a comment |
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
asked Nov 29 '18 at 2:44
user34104
1478
add a comment |
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
asked Nov 29 '18 at 2:44
user34104
1478
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
asked Nov 29 '18 at 2:44
user34104
1478
asked Nov 29 '18 at 2:44
user34104
1478
asked Nov 29 '18 at 2:44
user34104
1478
asked Nov 29 '18 at 2:44
user34104
1478
asked Nov 29 '18 at 2:44
user34104
1478
1478
add a comment |
add a comment |
1 Answer
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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,
$$
G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
$$
answered Nov 29 '18 at 21:47
munchhausen
79416
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
oldest
votes
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,
$$
G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
$$
answered Nov 29 '18 at 21:47
munchhausen
79416
add a comment |
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,
$$
G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
$$
answered Nov 29 '18 at 21:47
munchhausen
79416
add a comment |
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,
$$
G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
$$
answered Nov 29 '18 at 21:47
munchhausen
79416
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,
$$
G(q)=sum_n t_nq^n=t_0+t_1q+t_2q^2+sum_{ngeq 3}(f_{n-1}+f_{n-3})q^n=1+2q+3q^2+q(F(q)-f_0-f_1q)+q^3F(q)=1+q+q^2+(q+q^3)F(q).
$$
answered Nov 29 '18 at 21:47
munchhausen
79416
answered Nov 29 '18 at 21:47
munchhausen
79416
answered Nov 29 '18 at 21:47
munchhausen
79416
answered Nov 29 '18 at 21:47
munchhausen
79416
79416
add a comment |
add a comment |
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