How many integer r-tuples are there such that sum of the absolute values of their entries is less than or...












0












$begingroup$


How many r-tuples are there such that sum of absolute values of entries is less than or equal to $n$?



That is, what is the cardinality of the set $ {(x_1,...,x_r): x_i in Z text{ and }mid x_1mid+ ... + mid x_r mid leq n}$?



This should give me the growth function of the group $Z^r$ under the generating set $S={e_1,...,e_r}$ where $ e_i={0, ...,0, 1,0,...,0}$ with $1$ in the $i$-th position. It's known that the answer to this question is: $sum_{k=0}^{r}2^k {r choose k}{n choose k}$. I'm trying to figure it out for myself.




Attempt:
I have been able to figure out that the cardinality of $ {(x_1,...,x_r): x_i in Z^+ text{ and }x_1+ ... + x_r leq n}$ is $n choose k$.



Also I calculated cardinality of $ {(x_1,...,x_k): x_i in Z^+ text{ and } x_1+ ... + x_k = n}$ to be $n-1 choose k-1$ via the stars and bars method.



However, I am a little stuck on combining these results (or otherwise) to figure out the question with the absolute values.






combinatorics discrete-mathematics combinations subgroup-growth







share|cite|improve this question











$endgroup$












  • $begingroup$
    Look at the alternative problem: In how many ways can you divide $n$ $1$'s with $r-1$ separators? Then do it for $1, 2, .., n$. And don't forget about negative numbers.
    $endgroup$
    – EuxhenH
    Jan 10 at 19:46












  • $begingroup$
    @coffeemath you're right, in that case, which is my first step in my attempt, there should be the restriction that the $x_i$ are positive integers. I will reflect that.
    $endgroup$
    – Mike
    Jan 10 at 19:55










  • $begingroup$
    @coffeemath lol, it is stars and bars, I edited it.
    $endgroup$
    – Mike
    Jan 10 at 20:15










  • $begingroup$
    @RossMillikan Thanks for the catch, I converted everything to $k$'s
    $endgroup$
    – Mike
    Jan 10 at 20:16
















0












$begingroup$


How many r-tuples are there such that sum of absolute values of entries is less than or equal to $n$?



That is, what is the cardinality of the set $ {(x_1,...,x_r): x_i in Z text{ and }mid x_1mid+ ... + mid x_r mid leq n}$?



This should give me the growth function of the group $Z^r$ under the generating set $S={e_1,...,e_r}$ where $ e_i={0, ...,0, 1,0,...,0}$ with $1$ in the $i$-th position. It's known that the answer to this question is: $sum_{k=0}^{r}2^k {r choose k}{n choose k}$. I'm trying to figure it out for myself.




Attempt:
I have been able to figure out that the cardinality of $ {(x_1,...,x_r): x_i in Z^+ text{ and }x_1+ ... + x_r leq n}$ is $n choose k$.



Also I calculated cardinality of $ {(x_1,...,x_k): x_i in Z^+ text{ and } x_1+ ... + x_k = n}$ to be $n-1 choose k-1$ via the stars and bars method.



However, I am a little stuck on combining these results (or otherwise) to figure out the question with the absolute values.






combinatorics discrete-mathematics combinations subgroup-growth







share|cite|improve this question











$endgroup$












  • $begingroup$
    Look at the alternative problem: In how many ways can you divide $n$ $1$'s with $r-1$ separators? Then do it for $1, 2, .., n$. And don't forget about negative numbers.
    $endgroup$
    – EuxhenH
    Jan 10 at 19:46












  • $begingroup$
    @coffeemath you're right, in that case, which is my first step in my attempt, there should be the restriction that the $x_i$ are positive integers. I will reflect that.
    $endgroup$
    – Mike
    Jan 10 at 19:55










  • $begingroup$
    @coffeemath lol, it is stars and bars, I edited it.
    $endgroup$
    – Mike
    Jan 10 at 20:15










  • $begingroup$
    @RossMillikan Thanks for the catch, I converted everything to $k$'s
    $endgroup$
    – Mike
    Jan 10 at 20:16














0












0








0


1



$begingroup$


How many r-tuples are there such that sum of absolute values of entries is less than or equal to $n$?



That is, what is the cardinality of the set $ {(x_1,...,x_r): x_i in Z text{ and }mid x_1mid+ ... + mid x_r mid leq n}$?



This should give me the growth function of the group $Z^r$ under the generating set $S={e_1,...,e_r}$ where $ e_i={0, ...,0, 1,0,...,0}$ with $1$ in the $i$-th position. It's known that the answer to this question is: $sum_{k=0}^{r}2^k {r choose k}{n choose k}$. I'm trying to figure it out for myself.




Attempt:
I have been able to figure out that the cardinality of $ {(x_1,...,x_r): x_i in Z^+ text{ and }x_1+ ... + x_r leq n}$ is $n choose k$.



Also I calculated cardinality of $ {(x_1,...,x_k): x_i in Z^+ text{ and } x_1+ ... + x_k = n}$ to be $n-1 choose k-1$ via the stars and bars method.



However, I am a little stuck on combining these results (or otherwise) to figure out the question with the absolute values.






combinatorics discrete-mathematics combinations subgroup-growth







share|cite|improve this question











$endgroup$




How many r-tuples are there such that sum of absolute values of entries is less than or equal to $n$?



That is, what is the cardinality of the set $ {(x_1,...,x_r): x_i in Z text{ and }mid x_1mid+ ... + mid x_r mid leq n}$?



This should give me the growth function of the group $Z^r$ under the generating set $S={e_1,...,e_r}$ where $ e_i={0, ...,0, 1,0,...,0}$ with $1$ in the $i$-th position. It's known that the answer to this question is: $sum_{k=0}^{r}2^k {r choose k}{n choose k}$. I'm trying to figure it out for myself.




Attempt:
I have been able to figure out that the cardinality of $ {(x_1,...,x_r): x_i in Z^+ text{ and }x_1+ ... + x_r leq n}$ is $n choose k$.



Also I calculated cardinality of $ {(x_1,...,x_k): x_i in Z^+ text{ and } x_1+ ... + x_k = n}$ to be $n-1 choose k-1$ via the stars and bars method.



However, I am a little stuck on combining these results (or otherwise) to figure out the question with the absolute values.






combinatorics discrete-mathematics combinations subgroup-growth




combinatorics discrete-mathematics combinations subgroup-growth






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 20:14







Mike

















asked Jan 10 at 19:22









MikeMike

700414




700414












  • $begingroup$
    Look at the alternative problem: In how many ways can you divide $n$ $1$'s with $r-1$ separators? Then do it for $1, 2, .., n$. And don't forget about negative numbers.
    $endgroup$
    – EuxhenH
    Jan 10 at 19:46












  • $begingroup$
    @coffeemath you're right, in that case, which is my first step in my attempt, there should be the restriction that the $x_i$ are positive integers. I will reflect that.
    $endgroup$
    – Mike
    Jan 10 at 19:55










  • $begingroup$
    @coffeemath lol, it is stars and bars, I edited it.
    $endgroup$
    – Mike
    Jan 10 at 20:15










  • $begingroup$
    @RossMillikan Thanks for the catch, I converted everything to $k$'s
    $endgroup$
    – Mike
    Jan 10 at 20:16


















  • $begingroup$
    Look at the alternative problem: In how many ways can you divide $n$ $1$'s with $r-1$ separators? Then do it for $1, 2, .., n$. And don't forget about negative numbers.
    $endgroup$
    – EuxhenH
    Jan 10 at 19:46












  • $begingroup$
    @coffeemath you're right, in that case, which is my first step in my attempt, there should be the restriction that the $x_i$ are positive integers. I will reflect that.
    $endgroup$
    – Mike
    Jan 10 at 19:55










  • $begingroup$
    @coffeemath lol, it is stars and bars, I edited it.
    $endgroup$
    – Mike
    Jan 10 at 20:15










  • $begingroup$
    @RossMillikan Thanks for the catch, I converted everything to $k$'s
    $endgroup$
    – Mike
    Jan 10 at 20:16
















$begingroup$
Look at the alternative problem: In how many ways can you divide $n$ $1$'s with $r-1$ separators? Then do it for $1, 2, .., n$. And don't forget about negative numbers.
$endgroup$
– EuxhenH
Jan 10 at 19:46






$begingroup$
Look at the alternative problem: In how many ways can you divide $n$ $1$'s with $r-1$ separators? Then do it for $1, 2, .., n$. And don't forget about negative numbers.
$endgroup$
– EuxhenH
Jan 10 at 19:46














$begingroup$
@coffeemath you're right, in that case, which is my first step in my attempt, there should be the restriction that the $x_i$ are positive integers. I will reflect that.
$endgroup$
– Mike
Jan 10 at 19:55




$begingroup$
@coffeemath you're right, in that case, which is my first step in my attempt, there should be the restriction that the $x_i$ are positive integers. I will reflect that.
$endgroup$
– Mike
Jan 10 at 19:55












$begingroup$
@coffeemath lol, it is stars and bars, I edited it.
$endgroup$
– Mike
Jan 10 at 20:15




$begingroup$
@coffeemath lol, it is stars and bars, I edited it.
$endgroup$
– Mike
Jan 10 at 20:15












$begingroup$
@RossMillikan Thanks for the catch, I converted everything to $k$'s
$endgroup$
– Mike
Jan 10 at 20:16




$begingroup$
@RossMillikan Thanks for the catch, I converted everything to $k$'s
$endgroup$
– Mike
Jan 10 at 20:16










1 Answer
1






active

oldest

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1












$begingroup$

Start from your result that the number of ways to sum $k$ positive numbers to $n$ or less is $n choose k$.



To get the number of ways to sum $k$ positive numbers and $r-k$ zeros to get $n$ or less you choose the positions of the zeros in $r choose k$ ways then choose the positive numbers in $n choose k$ ways, so the number of ways to sum $k$ positives and $r-k$ zeros to $n$ or less is ${r choose k}{n choose k}$.



Because of your absolute values we can choose the sign of the nonzero numbers in $2^k$ ways, so the number of ways to sum the absolute values of $k$ nonzero numbers and $r-k$ zeros is $2^k{r choose k}{n choose k}$.



Now we just sum over $k$ from $0$ to $r$, getting the desired result $$sum_{k=0}^r2^k{r choose k}{n choose k}$$



Added: Alpha gives a closed form using a hypergometric function
$$_2F_1(-n,-r;1;2)$$






share|cite|improve this answer











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    $begingroup$

    Start from your result that the number of ways to sum $k$ positive numbers to $n$ or less is $n choose k$.



    To get the number of ways to sum $k$ positive numbers and $r-k$ zeros to get $n$ or less you choose the positions of the zeros in $r choose k$ ways then choose the positive numbers in $n choose k$ ways, so the number of ways to sum $k$ positives and $r-k$ zeros to $n$ or less is ${r choose k}{n choose k}$.



    Because of your absolute values we can choose the sign of the nonzero numbers in $2^k$ ways, so the number of ways to sum the absolute values of $k$ nonzero numbers and $r-k$ zeros is $2^k{r choose k}{n choose k}$.



    Now we just sum over $k$ from $0$ to $r$, getting the desired result $$sum_{k=0}^r2^k{r choose k}{n choose k}$$



    Added: Alpha gives a closed form using a hypergometric function
    $$_2F_1(-n,-r;1;2)$$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Start from your result that the number of ways to sum $k$ positive numbers to $n$ or less is $n choose k$.



      To get the number of ways to sum $k$ positive numbers and $r-k$ zeros to get $n$ or less you choose the positions of the zeros in $r choose k$ ways then choose the positive numbers in $n choose k$ ways, so the number of ways to sum $k$ positives and $r-k$ zeros to $n$ or less is ${r choose k}{n choose k}$.



      Because of your absolute values we can choose the sign of the nonzero numbers in $2^k$ ways, so the number of ways to sum the absolute values of $k$ nonzero numbers and $r-k$ zeros is $2^k{r choose k}{n choose k}$.



      Now we just sum over $k$ from $0$ to $r$, getting the desired result $$sum_{k=0}^r2^k{r choose k}{n choose k}$$



      Added: Alpha gives a closed form using a hypergometric function
      $$_2F_1(-n,-r;1;2)$$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Start from your result that the number of ways to sum $k$ positive numbers to $n$ or less is $n choose k$.



        To get the number of ways to sum $k$ positive numbers and $r-k$ zeros to get $n$ or less you choose the positions of the zeros in $r choose k$ ways then choose the positive numbers in $n choose k$ ways, so the number of ways to sum $k$ positives and $r-k$ zeros to $n$ or less is ${r choose k}{n choose k}$.



        Because of your absolute values we can choose the sign of the nonzero numbers in $2^k$ ways, so the number of ways to sum the absolute values of $k$ nonzero numbers and $r-k$ zeros is $2^k{r choose k}{n choose k}$.



        Now we just sum over $k$ from $0$ to $r$, getting the desired result $$sum_{k=0}^r2^k{r choose k}{n choose k}$$



        Added: Alpha gives a closed form using a hypergometric function
        $$_2F_1(-n,-r;1;2)$$






        share|cite|improve this answer











        $endgroup$



        Start from your result that the number of ways to sum $k$ positive numbers to $n$ or less is $n choose k$.



        To get the number of ways to sum $k$ positive numbers and $r-k$ zeros to get $n$ or less you choose the positions of the zeros in $r choose k$ ways then choose the positive numbers in $n choose k$ ways, so the number of ways to sum $k$ positives and $r-k$ zeros to $n$ or less is ${r choose k}{n choose k}$.



        Because of your absolute values we can choose the sign of the nonzero numbers in $2^k$ ways, so the number of ways to sum the absolute values of $k$ nonzero numbers and $r-k$ zeros is $2^k{r choose k}{n choose k}$.



        Now we just sum over $k$ from $0$ to $r$, getting the desired result $$sum_{k=0}^r2^k{r choose k}{n choose k}$$



        Added: Alpha gives a closed form using a hypergometric function
        $$_2F_1(-n,-r;1;2)$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 12 at 6:52

























        answered Jan 10 at 20:23









        Ross MillikanRoss Millikan

        293k23197371




        293k23197371






























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