In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{and}; -1s?$












3












$begingroup$


In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{or}; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.



Edit:

e.g., if $n=3$
$$begin{aligned}0&=;;;0+0+0,\ 0&=;;;1-1+0,\ 0&=;;;1+0-1,\ 0&=;;;0+1-1,\ 0&=-1+1+0,\ 0&=-1+0+1,\ 0&=;;;0-1+1.end{aligned}$$
So for $n=3;$ there are $7$ ways.






combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Set up a recurrence perhaps.
    $endgroup$
    – paw88789
    Dec 29 '18 at 16:35












  • $begingroup$
    do you mean n each or a total of n?
    $endgroup$
    – player100
    Dec 29 '18 at 16:45










  • $begingroup$
    Sorry I meant that in total, the number of 0's, 1's and -1's is n
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 16:59










  • $begingroup$
    @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:11










  • $begingroup$
    I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:25


















3












$begingroup$


In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{or}; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.



Edit:

e.g., if $n=3$
$$begin{aligned}0&=;;;0+0+0,\ 0&=;;;1-1+0,\ 0&=;;;1+0-1,\ 0&=;;;0+1-1,\ 0&=-1+1+0,\ 0&=-1+0+1,\ 0&=;;;0-1+1.end{aligned}$$
So for $n=3;$ there are $7$ ways.






combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Set up a recurrence perhaps.
    $endgroup$
    – paw88789
    Dec 29 '18 at 16:35












  • $begingroup$
    do you mean n each or a total of n?
    $endgroup$
    – player100
    Dec 29 '18 at 16:45










  • $begingroup$
    Sorry I meant that in total, the number of 0's, 1's and -1's is n
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 16:59










  • $begingroup$
    @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:11










  • $begingroup$
    I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:25
















3












3








3


1



$begingroup$


In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{or}; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.



Edit:

e.g., if $n=3$
$$begin{aligned}0&=;;;0+0+0,\ 0&=;;;1-1+0,\ 0&=;;;1+0-1,\ 0&=;;;0+1-1,\ 0&=-1+1+0,\ 0&=-1+0+1,\ 0&=;;;0-1+1.end{aligned}$$
So for $n=3;$ there are $7$ ways.






combinatorics







share|cite|improve this question











$endgroup$




In how many ways can I write $0$ as a sum of $n; 0s, 1s ;text{or}; -1s?$ (Taking the order into account).
I suspect there is no closed formula to express the result, but I'd like someone to confirm it, or deny it.



Edit:

e.g., if $n=3$
$$begin{aligned}0&=;;;0+0+0,\ 0&=;;;1-1+0,\ 0&=;;;1+0-1,\ 0&=;;;0+1-1,\ 0&=-1+1+0,\ 0&=-1+0+1,\ 0&=;;;0-1+1.end{aligned}$$
So for $n=3;$ there are $7$ ways.






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 22:01









user376343

3,9834829




3,9834829










asked Dec 29 '18 at 16:31









Lucio TanziniLucio Tanzini

368114




368114








  • 1




    $begingroup$
    Set up a recurrence perhaps.
    $endgroup$
    – paw88789
    Dec 29 '18 at 16:35












  • $begingroup$
    do you mean n each or a total of n?
    $endgroup$
    – player100
    Dec 29 '18 at 16:45










  • $begingroup$
    Sorry I meant that in total, the number of 0's, 1's and -1's is n
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 16:59










  • $begingroup$
    @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:11










  • $begingroup$
    I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:25
















  • 1




    $begingroup$
    Set up a recurrence perhaps.
    $endgroup$
    – paw88789
    Dec 29 '18 at 16:35












  • $begingroup$
    do you mean n each or a total of n?
    $endgroup$
    – player100
    Dec 29 '18 at 16:45










  • $begingroup$
    Sorry I meant that in total, the number of 0's, 1's and -1's is n
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 16:59










  • $begingroup$
    @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:11










  • $begingroup$
    I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
    $endgroup$
    – timtfj
    Dec 29 '18 at 17:25










1




1




$begingroup$
Set up a recurrence perhaps.
$endgroup$
– paw88789
Dec 29 '18 at 16:35






$begingroup$
Set up a recurrence perhaps.
$endgroup$
– paw88789
Dec 29 '18 at 16:35














$begingroup$
do you mean n each or a total of n?
$endgroup$
– player100
Dec 29 '18 at 16:45




$begingroup$
do you mean n each or a total of n?
$endgroup$
– player100
Dec 29 '18 at 16:45












$begingroup$
Sorry I meant that in total, the number of 0's, 1's and -1's is n
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 16:59




$begingroup$
Sorry I meant that in total, the number of 0's, 1's and -1's is n
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 16:59












$begingroup$
@MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
$endgroup$
– timtfj
Dec 29 '18 at 17:11




$begingroup$
@MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
$endgroup$
– timtfj
Dec 29 '18 at 17:11












$begingroup$
I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
$endgroup$
– timtfj
Dec 29 '18 at 17:25






$begingroup$
I'd start here: for each $n$, we can choose how many times $0$ occurs ,and how many times $1$ occurs. $-1$ then has to match $1$. So if $0$ is used $r$ times and $1$ is used $s$ times, $r=n-2s$.Then for each choice of $r$ and $s$ you've got a defined permutation problem, with $r, s$ and $s$ items..
$endgroup$
– timtfj
Dec 29 '18 at 17:25












2 Answers
2






active

oldest

votes


















4












$begingroup$

$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43



















0












$begingroup$

The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43
















4












$begingroup$

$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43














4












4








4





$begingroup$

$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.






share|cite|improve this answer











$endgroup$



$s(n) = sum_{k=0}^{n/2} binom{n}{n-2k} cdot binom{2k}{k}$ The first term is the number of ways to arrange the zeroes, and then the second term arranges the parities. Now, we can simplify further:



$s(n) = sum_{k=0}^{n/2} frac{2k!}{2k!} cdot frac{n!}{(n-2k)!(k!)(k!)} = sum_{k=0}^{n/2} frac{n!}{(n-2k)!(k!)(k!)}$



Edit: put a few values into the OIES and came across trinomial coeffecients. In particular, s(n) is the n-th central trinomial coefficient, which has several closed forms, which you can find in the second link.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 21:48

























answered Dec 29 '18 at 18:53









Zachary HunterZachary Hunter

1,065314




1,065314












  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43


















  • $begingroup$
    This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    $endgroup$
    – timtfj
    Dec 29 '18 at 19:07










  • $begingroup$
    Whoops yeah I forgot a term
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:12










  • $begingroup$
    Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 19:31










  • $begingroup$
    @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 19:37










  • $begingroup$
    Update: found closed form.
    $endgroup$
    – Zachary Hunter
    Dec 29 '18 at 20:43
















$begingroup$
This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
$endgroup$
– timtfj
Dec 29 '18 at 19:07




$begingroup$
This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
$endgroup$
– timtfj
Dec 29 '18 at 19:07












$begingroup$
Whoops yeah I forgot a term
$endgroup$
– Zachary Hunter
Dec 29 '18 at 19:12




$begingroup$
Whoops yeah I forgot a term
$endgroup$
– Zachary Hunter
Dec 29 '18 at 19:12












$begingroup$
Yes, I got the same formula. Do you think It can be simplified into a closed formula?
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 19:31




$begingroup$
Yes, I got the same formula. Do you think It can be simplified into a closed formula?
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 19:31












$begingroup$
@LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
$endgroup$
– Zachary Hunter
Dec 29 '18 at 19:37




$begingroup$
@LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
$endgroup$
– Zachary Hunter
Dec 29 '18 at 19:37












$begingroup$
Update: found closed form.
$endgroup$
– Zachary Hunter
Dec 29 '18 at 20:43




$begingroup$
Update: found closed form.
$endgroup$
– Zachary Hunter
Dec 29 '18 at 20:43











0












$begingroup$

The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34
















0












$begingroup$

The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34














0












0








0





$begingroup$

The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.






share|cite|improve this answer











$endgroup$



The number of $0s, 1s$ and $-1s$ possible for a particular $n$ can be seen by the number of solutions to:



$$2p+q=n| p,q in Bbb Z^+$$



This can be done in $frac{n+1}{2}$ ways for odd $n$ and $frac{n+2}{2}$ ways for even $n$



You'll just need to account for positioning after this.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 17:16

























answered Dec 29 '18 at 16:55









Rhys HughesRhys Hughes

7,1051630




7,1051630












  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34


















  • $begingroup$
    Yes, but I'm afraid positioning is the main problem Indeed
    $endgroup$
    – Lucio Tanzini
    Dec 29 '18 at 17:34
















$begingroup$
Yes, but I'm afraid positioning is the main problem Indeed
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 17:34




$begingroup$
Yes, but I'm afraid positioning is the main problem Indeed
$endgroup$
– Lucio Tanzini
Dec 29 '18 at 17:34


















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