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Prove that $sum_n t_n x^n = fracx^k(1-x^2)^k(1-x) $

4

1

$begingroup$

Let $t_n$ be a number of sequences
$$ 1 le a_1 < a_2 < ... < a_k le n $$
such that $a_2i$ is even and $a_2i+1$ is odd.

Prove that $$sum_n t_n x^n = fracx^k(1-x^2)^k(1-x) $$

My attempt. I want to use there enumerators and I have 4 cases.

1. both $n$ and $k$ are even (other cases will be analogical)

Then enumerator will be $$(x+x^3+...+x^n-1)^k/2 (1+x^2+...+x^n)^k/2 = left(xcdot frac1-x^fracn-121-x^2 cdot frac1-x^fracn21-x^2 right)^k/2$$
But how can I get from there my thesis?

combinatorics discrete-mathematics generating-functions

share|cite|improve this question

edited Apr 1 at 9:40

Robert Z

102k1072145

asked Apr 1 at 8:45

trolleytrolley

17610

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  From the definition of $t_n$ it seems like only $a_i = i$ satisfies this?
  $endgroup$
  – George Dewhirst
  Apr 1 at 8:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgeDewhirst edited
  $endgroup$
  – trolley
  Apr 1 at 8:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why would this question on which the asker has visibly been working would be closed : it is absurd !
  $endgroup$
  – Jean Marie
  Apr 1 at 9:00
  

  
  
  
  

add a comment | 

4

1

$begingroup$

Let $t_n$ be a number of sequences
$$ 1 le a_1 < a_2 < ... < a_k le n $$
such that $a_2i$ is even and $a_2i+1$ is odd.

Prove that $$sum_n t_n x^n = fracx^k(1-x^2)^k(1-x) $$

My attempt. I want to use there enumerators and I have 4 cases.

1. both $n$ and $k$ are even (other cases will be analogical)

Then enumerator will be $$(x+x^3+...+x^n-1)^k/2 (1+x^2+...+x^n)^k/2 = left(xcdot frac1-x^fracn-121-x^2 cdot frac1-x^fracn21-x^2 right)^k/2$$
But how can I get from there my thesis?

combinatorics discrete-mathematics generating-functions

share|cite|improve this question

edited Apr 1 at 9:40

Robert Z

102k1072145

asked Apr 1 at 8:45

trolleytrolley

17610

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  From the definition of $t_n$ it seems like only $a_i = i$ satisfies this?
  $endgroup$
  – George Dewhirst
  Apr 1 at 8:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgeDewhirst edited
  $endgroup$
  – trolley
  Apr 1 at 8:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why would this question on which the asker has visibly been working would be closed : it is absurd !
  $endgroup$
  – Jean Marie
  Apr 1 at 9:00
  

  
  
  
  

add a comment | 

$begingroup$

Let $t_n$ be a number of sequences
$$ 1 le a_1 < a_2 < ... < a_k le n $$
such that $a_2i$ is even and $a_2i+1$ is odd.

Prove that $$sum_n t_n x^n = fracx^k(1-x^2)^k(1-x) $$

My attempt. I want to use there enumerators and I have 4 cases.

1. both $n$ and $k$ are even (other cases will be analogical)

Then enumerator will be $$(x+x^3+...+x^n-1)^k/2 (1+x^2+...+x^n)^k/2 = left(xcdot frac1-x^fracn-121-x^2 cdot frac1-x^fracn21-x^2 right)^k/2$$
But how can I get from there my thesis?

combinatorics discrete-mathematics generating-functions

share|cite|improve this question

edited Apr 1 at 9:40

Robert Z

102k1072145

asked Apr 1 at 8:45

trolleytrolley

17610

$endgroup$

share|cite|improve this question

edited Apr 1 at 9:40

Robert Z

102k1072145

asked Apr 1 at 8:45

trolleytrolley

17610

share|cite|improve this question

edited Apr 1 at 9:40

Robert Z

102k1072145

asked Apr 1 at 8:45

trolleytrolley

17610

edited Apr 1 at 9:40

Robert Z

102k1072145

edited Apr 1 at 9:40

Robert Z

102k1072145

asked Apr 1 at 8:45

trolleytrolley

17610

asked Apr 1 at 8:45

trolleytrolley

17610

1 Answer
1

active

oldest

votes

2

$begingroup$

We have that
$$f(x):=fracx^k(1-x^2)^k(1-x)=frac11-xleft(fracx1-x^2right)^k
=frac11-xleft(sum_j=0^inftyx^2j+1right)^k.$$
Therefore the coefficient $t_n=[x^n]f(x)$ is the number of ways we can write a positive integer $leq n$ as the sum of $k$ odd numbers.
Now note that $a_1,a_2-a_1,dots,a_k-a_k-1$ are $k$ odd numbers whose sum is $a_k$ which is $leq n$ and such $k$ odd numbers determine in unique way a sequence $ 1 le a_1 < a_2 < ... < a_k le n $ such that $a_2i$ is even and $a_2i+1$ is odd.

share|cite|improve this answer

edited Apr 1 at 17:20

answered Apr 1 at 9:33

Robert ZRobert Z

102k1072145

$endgroup$

add a comment | 

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2

$begingroup$

We have that
$$f(x):=fracx^k(1-x^2)^k(1-x)=frac11-xleft(fracx1-x^2right)^k
=frac11-xleft(sum_j=0^inftyx^2j+1right)^k.$$
Therefore the coefficient $t_n=[x^n]f(x)$ is the number of ways we can write a positive integer $leq n$ as the sum of $k$ odd numbers.
Now note that $a_1,a_2-a_1,dots,a_k-a_k-1$ are $k$ odd numbers whose sum is $a_k$ which is $leq n$ and such $k$ odd numbers determine in unique way a sequence $ 1 le a_1 < a_2 < ... < a_k le n $ such that $a_2i$ is even and $a_2i+1$ is odd.

share|cite|improve this answer

edited Apr 1 at 17:20

answered Apr 1 at 9:33

Robert ZRobert Z

102k1072145

$endgroup$

add a comment | 

2

$begingroup$

We have that
$$f(x):=fracx^k(1-x^2)^k(1-x)=frac11-xleft(fracx1-x^2right)^k
=frac11-xleft(sum_j=0^inftyx^2j+1right)^k.$$
Therefore the coefficient $t_n=[x^n]f(x)$ is the number of ways we can write a positive integer $leq n$ as the sum of $k$ odd numbers.
Now note that $a_1,a_2-a_1,dots,a_k-a_k-1$ are $k$ odd numbers whose sum is $a_k$ which is $leq n$ and such $k$ odd numbers determine in unique way a sequence $ 1 le a_1 < a_2 < ... < a_k le n $ such that $a_2i$ is even and $a_2i+1$ is odd.

share|cite|improve this answer

edited Apr 1 at 17:20

answered Apr 1 at 9:33

Robert ZRobert Z

102k1072145

$endgroup$

add a comment | 

$begingroup$

We have that
$$f(x):=fracx^k(1-x^2)^k(1-x)=frac11-xleft(fracx1-x^2right)^k
=frac11-xleft(sum_j=0^inftyx^2j+1right)^k.$$
Therefore the coefficient $t_n=[x^n]f(x)$ is the number of ways we can write a positive integer $leq n$ as the sum of $k$ odd numbers.
Now note that $a_1,a_2-a_1,dots,a_k-a_k-1$ are $k$ odd numbers whose sum is $a_k$ which is $leq n$ and such $k$ odd numbers determine in unique way a sequence $ 1 le a_1 < a_2 < ... < a_k le n $ such that $a_2i$ is even and $a_2i+1$ is odd.

share|cite|improve this answer

edited Apr 1 at 17:20

answered Apr 1 at 9:33

Robert ZRobert Z

102k1072145

$endgroup$

share|cite|improve this answer

edited Apr 1 at 17:20

answered Apr 1 at 9:33

Robert ZRobert Z

102k1072145

answered Apr 1 at 9:33

Robert ZRobert Z

102k1072145

answered Apr 1 at 9:33

Robert ZRobert Z

102k1072145