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












3














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







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  • 1




    Set up a recurrence perhaps.
    – paw88789
    5 hours ago












  • do you mean n each or a total of n?
    – player100
    5 hours ago










  • Sorry I meant that in total, the number of 0's, 1's and -1's is n
    – Lucio Tanzini
    5 hours ago










  • @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    – timtfj
    5 hours ago










  • 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..
    – timtfj
    4 hours ago


















3














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




















  • 1




    Set up a recurrence perhaps.
    – paw88789
    5 hours ago












  • do you mean n each or a total of n?
    – player100
    5 hours ago










  • Sorry I meant that in total, the number of 0's, 1's and -1's is n
    – Lucio Tanzini
    5 hours ago










  • @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    – timtfj
    5 hours ago










  • 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..
    – timtfj
    4 hours ago
















3












3








3


2





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















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 14 mins ago









user376343

2,7982822




2,7982822










asked 5 hours ago









Lucio Tanzini

17514




17514








  • 1




    Set up a recurrence perhaps.
    – paw88789
    5 hours ago












  • do you mean n each or a total of n?
    – player100
    5 hours ago










  • Sorry I meant that in total, the number of 0's, 1's and -1's is n
    – Lucio Tanzini
    5 hours ago










  • @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    – timtfj
    5 hours ago










  • 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..
    – timtfj
    4 hours ago
















  • 1




    Set up a recurrence perhaps.
    – paw88789
    5 hours ago












  • do you mean n each or a total of n?
    – player100
    5 hours ago










  • Sorry I meant that in total, the number of 0's, 1's and -1's is n
    – Lucio Tanzini
    5 hours ago










  • @MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
    – timtfj
    5 hours ago










  • 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..
    – timtfj
    4 hours ago










1




1




Set up a recurrence perhaps.
– paw88789
5 hours ago






Set up a recurrence perhaps.
– paw88789
5 hours ago














do you mean n each or a total of n?
– player100
5 hours ago




do you mean n each or a total of n?
– player100
5 hours ago












Sorry I meant that in total, the number of 0's, 1's and -1's is n
– Lucio Tanzini
5 hours ago




Sorry I meant that in total, the number of 0's, 1's and -1's is n
– Lucio Tanzini
5 hours ago












@MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
– timtfj
5 hours ago




@MJD Thsnks for pointing out that I'd missed the $n$. I've deleted the answer!
– timtfj
5 hours ago












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..
– timtfj
4 hours ago






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..
– timtfj
4 hours ago












2 Answers
2






active

oldest

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3














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






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New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    – timtfj
    3 hours ago










  • Whoops yeah I forgot a term
    – Zachary Hunter
    3 hours ago










  • Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    – Lucio Tanzini
    2 hours ago










  • @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    – Zachary Hunter
    2 hours ago










  • Update: found closed form.
    – Zachary Hunter
    1 hour ago



















0














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























  • Yes, but I'm afraid positioning is the main problem Indeed
    – Lucio Tanzini
    4 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














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










New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    – timtfj
    3 hours ago










  • Whoops yeah I forgot a term
    – Zachary Hunter
    3 hours ago










  • Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    – Lucio Tanzini
    2 hours ago










  • @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    – Zachary Hunter
    2 hours ago










  • Update: found closed form.
    – Zachary Hunter
    1 hour ago
















3














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










New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    – timtfj
    3 hours ago










  • Whoops yeah I forgot a term
    – Zachary Hunter
    3 hours ago










  • Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    – Lucio Tanzini
    2 hours ago










  • @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    – Zachary Hunter
    2 hours ago










  • Update: found closed form.
    – Zachary Hunter
    1 hour ago














3












3








3






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










New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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










New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer








edited 27 mins ago





















New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 3 hours ago









Zachary Hunter

595




595




New contributor




Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    – timtfj
    3 hours ago










  • Whoops yeah I forgot a term
    – Zachary Hunter
    3 hours ago










  • Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    – Lucio Tanzini
    2 hours ago










  • @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    – Zachary Hunter
    2 hours ago










  • Update: found closed form.
    – Zachary Hunter
    1 hour ago


















  • This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
    – timtfj
    3 hours ago










  • Whoops yeah I forgot a term
    – Zachary Hunter
    3 hours ago










  • Yes, I got the same formula. Do you think It can be simplified into a closed formula?
    – Lucio Tanzini
    2 hours ago










  • @LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
    – Zachary Hunter
    2 hours ago










  • Update: found closed form.
    – Zachary Hunter
    1 hour ago
















This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
– timtfj
3 hours ago




This looks wrong to me. There are $k$ $1$'s, $k$ $(-1)$'s and $n-2k$ zeroes, so the denominator should be $(k!)^2(n-2k)!$.
– timtfj
3 hours ago












Whoops yeah I forgot a term
– Zachary Hunter
3 hours ago




Whoops yeah I forgot a term
– Zachary Hunter
3 hours ago












Yes, I got the same formula. Do you think It can be simplified into a closed formula?
– Lucio Tanzini
2 hours ago




Yes, I got the same formula. Do you think It can be simplified into a closed formula?
– Lucio Tanzini
2 hours ago












@LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
– Zachary Hunter
2 hours ago




@LucioTanzini nothing immediately comes to mind without getting my hands dirty, but I’m thinking on it.
– Zachary Hunter
2 hours ago












Update: found closed form.
– Zachary Hunter
1 hour ago




Update: found closed form.
– Zachary Hunter
1 hour ago











0














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























  • Yes, but I'm afraid positioning is the main problem Indeed
    – Lucio Tanzini
    4 hours ago
















0














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























  • Yes, but I'm afraid positioning is the main problem Indeed
    – Lucio Tanzini
    4 hours ago














0












0








0






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














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 4 hours ago

























answered 5 hours ago









Rhys Hughes

4,7961327




4,7961327












  • Yes, but I'm afraid positioning is the main problem Indeed
    – Lucio Tanzini
    4 hours ago


















  • Yes, but I'm afraid positioning is the main problem Indeed
    – Lucio Tanzini
    4 hours ago
















Yes, but I'm afraid positioning is the main problem Indeed
– Lucio Tanzini
4 hours ago




Yes, but I'm afraid positioning is the main problem Indeed
– Lucio Tanzini
4 hours ago


















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Sidney Franklin

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Questa voce sull'argomento registi statunitensi è solo un abbozzo . Contribuisci a migliorarla secondo le convenzioni di Wikipedia. Sidney Franklin, foto di Carl Van Vechten Oscar alla memoria Irving G. Thalberg 1943 Sidney Franklin , all'anagrafe Sidney Arnold Franklin , (San Francisco, 21 marzo 1893 – Santa Monica, 18 maggio 1972), è stato un regista, produttore cinematografico e attore statunitense. Il 20 giugno 1963, si sposò con l'attrice australiana Enid Bennett. Il matrimonio durò fino alla morte dell'attrice, il 14 maggio 1969 [1] Indice 1 Filmografia 1.1 Regista 1.2 Aiuto regia (parziale) 1.3 Produttore 1.4 Attore (parziale) 2 Note 3 Altri progetti 4 Collegamenti esterni Filmografia | Regista | The Baby , co-regia Chester M. Franklin (1915) The Rivals , co-regia Chester M. Franklin (1915) Little Dick's First Case , co-regia Chester M. Franklin (1915) Her Filmland Hero , co-...
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