Mixing
Find the recurrence formula!
$begingroup$
I have a sequence defined by recursion as follows:
$$begin{cases}x_0=a\ x_{n+1}=x_ncdot B^{x_n} end{cases}$$
where $a,B$ are fix natural numbers. Does anyone know how to find a recurrence formula for this?
I tried to write in a different way, and I figure out that another equivalent definition of the sequence could be
$$begin{cases}x_0=a\ x_{n+1}=acdot B^{x_0+cdots+x_n}end{cases}$$
Then I tried to use logarithms and differences, but really couldn't get to anything good.
sequences-and-series combinatorics number-theory recurrence-relations
asked Jan 17 at 16:07
Darío GDarío G
4,062613
$endgroup$
add a comment |
$begingroup$
I have a sequence defined by recursion as follows:
$$begin{cases}x_0=a\ x_{n+1}=x_ncdot B^{x_n} end{cases}$$
where $a,B$ are fix natural numbers. Does anyone know how to find a recurrence formula for this?
I tried to write in a different way, and I figure out that another equivalent definition of the sequence could be
$$begin{cases}x_0=a\ x_{n+1}=acdot B^{x_0+cdots+x_n}end{cases}$$
Then I tried to use logarithms and differences, but really couldn't get to anything good.
sequences-and-series combinatorics number-theory recurrence-relations
asked Jan 17 at 16:07
Darío GDarío G
4,062613
$endgroup$
$begingroup$
Why do you assume you can simplify this more?
$endgroup$
– Don Thousand
Jan 17 at 16:20
$begingroup$
Not assuming. Just wondering if there is a way to get a recurrence formula
$endgroup$
– Darío G
Jan 17 at 16:23
2
$begingroup$
Forget any hope.
$endgroup$
– Yves Daoust
Jan 17 at 18:25
add a comment |
$begingroup$
I have a sequence defined by recursion as follows:
$$begin{cases}x_0=a\ x_{n+1}=x_ncdot B^{x_n} end{cases}$$
where $a,B$ are fix natural numbers. Does anyone know how to find a recurrence formula for this?
I tried to write in a different way, and I figure out that another equivalent definition of the sequence could be
$$begin{cases}x_0=a\ x_{n+1}=acdot B^{x_0+cdots+x_n}end{cases}$$
Then I tried to use logarithms and differences, but really couldn't get to anything good.
sequences-and-series combinatorics number-theory recurrence-relations
asked Jan 17 at 16:07
Darío GDarío G
4,062613
$endgroup$
I have a sequence defined by recursion as follows:
$$begin{cases}x_0=a\ x_{n+1}=x_ncdot B^{x_n} end{cases}$$
where $a,B$ are fix natural numbers. Does anyone know how to find a recurrence formula for this?
I tried to write in a different way, and I figure out that another equivalent definition of the sequence could be
$$begin{cases}x_0=a\ x_{n+1}=acdot B^{x_0+cdots+x_n}end{cases}$$
Then I tried to use logarithms and differences, but really couldn't get to anything good.
sequences-and-series combinatorics number-theory recurrence-relations
sequences-and-series combinatorics number-theory recurrence-relations
asked Jan 17 at 16:07
Darío GDarío G
4,062613
asked Jan 17 at 16:07
Darío GDarío G
4,062613
asked Jan 17 at 16:07
Darío GDarío G
4,062613
asked Jan 17 at 16:07
Darío GDarío G
4,062613
asked Jan 17 at 16:07
Darío GDarío G
4,062613
4,062613
$begingroup$
Why do you assume you can simplify this more?
$endgroup$
– Don Thousand
Jan 17 at 16:20
$begingroup$
Not assuming. Just wondering if there is a way to get a recurrence formula
$endgroup$
– Darío G
Jan 17 at 16:23
2
$begingroup$
Forget any hope.
$endgroup$
– Yves Daoust
Jan 17 at 18:25
add a comment |
$begingroup$
Why do you assume you can simplify this more?
$endgroup$
– Don Thousand
Jan 17 at 16:20
$begingroup$
Not assuming. Just wondering if there is a way to get a recurrence formula
$endgroup$
– Darío G
Jan 17 at 16:23
2
$begingroup$
Forget any hope.
$endgroup$
– Yves Daoust
Jan 17 at 18:25
$begingroup$
Why do you assume you can simplify this more?
$endgroup$
– Don Thousand
Jan 17 at 16:20
$begingroup$
Why do you assume you can simplify this more?
$endgroup$
– Don Thousand
Jan 17 at 16:20
$begingroup$
Not assuming. Just wondering if there is a way to get a recurrence formula
$endgroup$
– Darío G
Jan 17 at 16:23
$begingroup$
Not assuming. Just wondering if there is a way to get a recurrence formula
$endgroup$
– Darío G
Jan 17 at 16:23
2
2
$begingroup$
Forget any hope.
$endgroup$
– Yves Daoust
Jan 17 at 18:25
$begingroup$
Forget any hope.
$endgroup$
– Yves Daoust
Jan 17 at 18:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You wanted to say : an explicit formula for $x_n$ I suppose ?
If you calculate the firsts terms :
$x_0 = a$
$x_1 = a B^a$
$x_2 = a B^a B^{aB^a}$
$x_3 = a B^a B^{aB^a} B^{a B^a B^{aB^a}}$
You can guess $x_n$ can be written in the form :
$x_n = B_0 B_1 .. B_n$
[EDIT : what I have written just after this was false]
EDIT : You have :
$x_n = B_0 B_1 .. B_n$
With $B_0 = a, B_n = B^{B_0 .. B_{n-1}}$
So you can understand the terms of $x_n$ using decreasing sequences of integers with first terms $leq n$.
For instance :
n = 2.
$x_2 = a B^a B^{aB^a}$
To (0) will be associated the first $a$ at the left.
To (1) is associated the tower $B^a$. To (1,0) is associated the $a$ in the tower $B^a$
To (2) is associated the tower $B^{aB^a}$. To $(2,0)$, the first $a$ at the left in the exponent of the tower $aB^a$. To $(2,1)$ the term $B^a$ in $aB^a$. To (2,1,0) the last exponent $a$.
In the general case, you have an algorithm to generate your expression of $x_n$ in function of $a$ and $B$.
- Put the $a$ on the first floor, associated to $(0)$, put the $n$ $B$ on the first floor, the base of the towers associated to the sequences of 1 term (1), (2), .. , (n).
- To fill the second floor, you look at the sequences $(k, l)$ with $0 leq l < k$. To each of these sequence, you put a B on the $k$-th tower if $lneq 0$, $a$ if $l=0$
- You continue like this, until the $n$-th floor ; then you have finished.
If you index $B$ by the sequences I said, you can also have a recursion formula.
$B_r = a$ if $r$ is a sequence ending by $0$, $B_r = B^{prod B_s}$
where the product is on the $s$, the sequences which has $r$ as initial segment and one term more than r, and $x_n = B_{(0)} .. B_{(n)}$
It is quite technical, but it enables to understand how the towers of $B$ and $a$ are distributed, and can be helpful if you want to have some asymptotics.
answered Jan 17 at 16:24
DLeMeurDLeMeur
3148
$endgroup$
$begingroup$
Of course, although this is not very useful for making future calculations
$endgroup$
– Darío G
Jan 17 at 16:30
$begingroup$
This kind of reasoning might be useful if you want to guess asymptotics of the sequence.
$endgroup$
– DLeMeur
Jan 17 at 16:39
add a comment |
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
You wanted to say : an explicit formula for $x_n$ I suppose ?
If you calculate the firsts terms :
$x_0 = a$
$x_1 = a B^a$
$x_2 = a B^a B^{aB^a}$
$x_3 = a B^a B^{aB^a} B^{a B^a B^{aB^a}}$
You can guess $x_n$ can be written in the form :
$x_n = B_0 B_1 .. B_n$
[EDIT : what I have written just after this was false]
EDIT : You have :
$x_n = B_0 B_1 .. B_n$
With $B_0 = a, B_n = B^{B_0 .. B_{n-1}}$
So you can understand the terms of $x_n$ using decreasing sequences of integers with first terms $leq n$.
For instance :
n = 2.
$x_2 = a B^a B^{aB^a}$
To (0) will be associated the first $a$ at the left.
To (1) is associated the tower $B^a$. To (1,0) is associated the $a$ in the tower $B^a$
To (2) is associated the tower $B^{aB^a}$. To $(2,0)$, the first $a$ at the left in the exponent of the tower $aB^a$. To $(2,1)$ the term $B^a$ in $aB^a$. To (2,1,0) the last exponent $a$.
In the general case, you have an algorithm to generate your expression of $x_n$ in function of $a$ and $B$.
- Put the $a$ on the first floor, associated to $(0)$, put the $n$ $B$ on the first floor, the base of the towers associated to the sequences of 1 term (1), (2), .. , (n).
- To fill the second floor, you look at the sequences $(k, l)$ with $0 leq l < k$. To each of these sequence, you put a B on the $k$-th tower if $lneq 0$, $a$ if $l=0$
- You continue like this, until the $n$-th floor ; then you have finished.
If you index $B$ by the sequences I said, you can also have a recursion formula.
$B_r = a$ if $r$ is a sequence ending by $0$, $B_r = B^{prod B_s}$
where the product is on the $s$, the sequences which has $r$ as initial segment and one term more than r, and $x_n = B_{(0)} .. B_{(n)}$
It is quite technical, but it enables to understand how the towers of $B$ and $a$ are distributed, and can be helpful if you want to have some asymptotics.
answered Jan 17 at 16:24
DLeMeurDLeMeur
3148
$endgroup$
$begingroup$
Of course, although this is not very useful for making future calculations
$endgroup$
– Darío G
Jan 17 at 16:30
$begingroup$
This kind of reasoning might be useful if you want to guess asymptotics of the sequence.
$endgroup$
– DLeMeur
Jan 17 at 16:39
add a comment |
$begingroup$
You wanted to say : an explicit formula for $x_n$ I suppose ?
If you calculate the firsts terms :
$x_0 = a$
$x_1 = a B^a$
$x_2 = a B^a B^{aB^a}$
$x_3 = a B^a B^{aB^a} B^{a B^a B^{aB^a}}$
You can guess $x_n$ can be written in the form :
$x_n = B_0 B_1 .. B_n$
[EDIT : what I have written just after this was false]
EDIT : You have :
$x_n = B_0 B_1 .. B_n$
With $B_0 = a, B_n = B^{B_0 .. B_{n-1}}$
So you can understand the terms of $x_n$ using decreasing sequences of integers with first terms $leq n$.
For instance :
n = 2.
$x_2 = a B^a B^{aB^a}$
To (0) will be associated the first $a$ at the left.
To (1) is associated the tower $B^a$. To (1,0) is associated the $a$ in the tower $B^a$
To (2) is associated the tower $B^{aB^a}$. To $(2,0)$, the first $a$ at the left in the exponent of the tower $aB^a$. To $(2,1)$ the term $B^a$ in $aB^a$. To (2,1,0) the last exponent $a$.
In the general case, you have an algorithm to generate your expression of $x_n$ in function of $a$ and $B$.
- Put the $a$ on the first floor, associated to $(0)$, put the $n$ $B$ on the first floor, the base of the towers associated to the sequences of 1 term (1), (2), .. , (n).
- To fill the second floor, you look at the sequences $(k, l)$ with $0 leq l < k$. To each of these sequence, you put a B on the $k$-th tower if $lneq 0$, $a$ if $l=0$
- You continue like this, until the $n$-th floor ; then you have finished.
If you index $B$ by the sequences I said, you can also have a recursion formula.
$B_r = a$ if $r$ is a sequence ending by $0$, $B_r = B^{prod B_s}$
where the product is on the $s$, the sequences which has $r$ as initial segment and one term more than r, and $x_n = B_{(0)} .. B_{(n)}$
It is quite technical, but it enables to understand how the towers of $B$ and $a$ are distributed, and can be helpful if you want to have some asymptotics.
answered Jan 17 at 16:24
DLeMeurDLeMeur
3148
$endgroup$
$begingroup$
Of course, although this is not very useful for making future calculations
$endgroup$
– Darío G
Jan 17 at 16:30
$begingroup$
This kind of reasoning might be useful if you want to guess asymptotics of the sequence.
$endgroup$
– DLeMeur
Jan 17 at 16:39
add a comment |
$begingroup$
You wanted to say : an explicit formula for $x_n$ I suppose ?
If you calculate the firsts terms :
$x_0 = a$
$x_1 = a B^a$
$x_2 = a B^a B^{aB^a}$
$x_3 = a B^a B^{aB^a} B^{a B^a B^{aB^a}}$
You can guess $x_n$ can be written in the form :
$x_n = B_0 B_1 .. B_n$
[EDIT : what I have written just after this was false]
EDIT : You have :
$x_n = B_0 B_1 .. B_n$
With $B_0 = a, B_n = B^{B_0 .. B_{n-1}}$
So you can understand the terms of $x_n$ using decreasing sequences of integers with first terms $leq n$.
For instance :
n = 2.
$x_2 = a B^a B^{aB^a}$
To (0) will be associated the first $a$ at the left.
To (1) is associated the tower $B^a$. To (1,0) is associated the $a$ in the tower $B^a$
To (2) is associated the tower $B^{aB^a}$. To $(2,0)$, the first $a$ at the left in the exponent of the tower $aB^a$. To $(2,1)$ the term $B^a$ in $aB^a$. To (2,1,0) the last exponent $a$.
In the general case, you have an algorithm to generate your expression of $x_n$ in function of $a$ and $B$.
- Put the $a$ on the first floor, associated to $(0)$, put the $n$ $B$ on the first floor, the base of the towers associated to the sequences of 1 term (1), (2), .. , (n).
- To fill the second floor, you look at the sequences $(k, l)$ with $0 leq l < k$. To each of these sequence, you put a B on the $k$-th tower if $lneq 0$, $a$ if $l=0$
- You continue like this, until the $n$-th floor ; then you have finished.
If you index $B$ by the sequences I said, you can also have a recursion formula.
$B_r = a$ if $r$ is a sequence ending by $0$, $B_r = B^{prod B_s}$
where the product is on the $s$, the sequences which has $r$ as initial segment and one term more than r, and $x_n = B_{(0)} .. B_{(n)}$
It is quite technical, but it enables to understand how the towers of $B$ and $a$ are distributed, and can be helpful if you want to have some asymptotics.
answered Jan 17 at 16:24
DLeMeurDLeMeur
3148
$endgroup$
You wanted to say : an explicit formula for $x_n$ I suppose ?
If you calculate the firsts terms :
$x_0 = a$
$x_1 = a B^a$
$x_2 = a B^a B^{aB^a}$
$x_3 = a B^a B^{aB^a} B^{a B^a B^{aB^a}}$
You can guess $x_n$ can be written in the form :
$x_n = B_0 B_1 .. B_n$
[EDIT : what I have written just after this was false]
EDIT : You have :
$x_n = B_0 B_1 .. B_n$
With $B_0 = a, B_n = B^{B_0 .. B_{n-1}}$
So you can understand the terms of $x_n$ using decreasing sequences of integers with first terms $leq n$.
For instance :
n = 2.
$x_2 = a B^a B^{aB^a}$
To (0) will be associated the first $a$ at the left.
To (1) is associated the tower $B^a$. To (1,0) is associated the $a$ in the tower $B^a$
To (2) is associated the tower $B^{aB^a}$. To $(2,0)$, the first $a$ at the left in the exponent of the tower $aB^a$. To $(2,1)$ the term $B^a$ in $aB^a$. To (2,1,0) the last exponent $a$.
In the general case, you have an algorithm to generate your expression of $x_n$ in function of $a$ and $B$.
- Put the $a$ on the first floor, associated to $(0)$, put the $n$ $B$ on the first floor, the base of the towers associated to the sequences of 1 term (1), (2), .. , (n).
- To fill the second floor, you look at the sequences $(k, l)$ with $0 leq l < k$. To each of these sequence, you put a B on the $k$-th tower if $lneq 0$, $a$ if $l=0$
- You continue like this, until the $n$-th floor ; then you have finished.
If you index $B$ by the sequences I said, you can also have a recursion formula.
$B_r = a$ if $r$ is a sequence ending by $0$, $B_r = B^{prod B_s}$
where the product is on the $s$, the sequences which has $r$ as initial segment and one term more than r, and $x_n = B_{(0)} .. B_{(n)}$
It is quite technical, but it enables to understand how the towers of $B$ and $a$ are distributed, and can be helpful if you want to have some asymptotics.
answered Jan 17 at 16:24
DLeMeurDLeMeur
3148
edited Jan 17 at 18:21
answered Jan 17 at 16:24
DLeMeurDLeMeur
3148
answered Jan 17 at 16:24
DLeMeurDLeMeur
3148
answered Jan 17 at 16:24
DLeMeurDLeMeur
3148
3148
$begingroup$
Of course, although this is not very useful for making future calculations
$endgroup$
– Darío G
Jan 17 at 16:30
$begingroup$
This kind of reasoning might be useful if you want to guess asymptotics of the sequence.
$endgroup$
– DLeMeur
Jan 17 at 16:39
add a comment |
$begingroup$
Of course, although this is not very useful for making future calculations
$endgroup$
– Darío G
Jan 17 at 16:30
$begingroup$
This kind of reasoning might be useful if you want to guess asymptotics of the sequence.
$endgroup$
– DLeMeur
Jan 17 at 16:39
$begingroup$
Of course, although this is not very useful for making future calculations
$endgroup$
– Darío G
Jan 17 at 16:30
$begingroup$
Of course, although this is not very useful for making future calculations
$endgroup$
– Darío G
Jan 17 at 16:30
$begingroup$
This kind of reasoning might be useful if you want to guess asymptotics of the sequence.
$endgroup$
– DLeMeur
Jan 17 at 16:39
$begingroup$
This kind of reasoning might be useful if you want to guess asymptotics of the sequence.
$endgroup$
– DLeMeur
Jan 17 at 16:39
add a comment |
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$begingroup$
Why do you assume you can simplify this more?
$endgroup$
– Don Thousand
Jan 17 at 16:20
$begingroup$
Not assuming. Just wondering if there is a way to get a recurrence formula
$endgroup$
– Darío G
Jan 17 at 16:23
2
$begingroup$
Forget any hope.
$endgroup$
– Yves Daoust
Jan 17 at 18:25