Find a good formulae
$begingroup$
We know that equation $$s_1+s_2+s_3=n-1 quad mbox{$s_1,s_2,s_3$}geq 1$$
has $binom{n-2}{2}$ solution.
I want to find any good formulae for the following form :
$$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=?$$
where, $s_0=1$ and each $(s_1,s_2,s_3)$ is the solution of above equaiton.
- My all attempts have failed.
$textbf{Edit}:$ I also note that following:
$$n=4Longrightarrow 1$$
$$n=5Longrightarrow 5$$
$$n=6Longrightarrow 18$$
$$n=7Longrightarrow 53$$
$$n=8Longrightarrow 169$$
$$n=9Longrightarrow 502$$
combinatorics discrete-mathematics combinations
asked Dec 10 '18 at 13:28
1Spectre11Spectre1
999
$endgroup$
add a comment |
$begingroup$
We know that equation $$s_1+s_2+s_3=n-1 quad mbox{$s_1,s_2,s_3$}geq 1$$
has $binom{n-2}{2}$ solution.
I want to find any good formulae for the following form :
$$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=?$$
where, $s_0=1$ and each $(s_1,s_2,s_3)$ is the solution of above equaiton.
- My all attempts have failed.
$textbf{Edit}:$ I also note that following:
$$n=4Longrightarrow 1$$
$$n=5Longrightarrow 5$$
$$n=6Longrightarrow 18$$
$$n=7Longrightarrow 53$$
$$n=8Longrightarrow 169$$
$$n=9Longrightarrow 502$$
combinatorics discrete-mathematics combinations
asked Dec 10 '18 at 13:28
1Spectre11Spectre1
999
$endgroup$
3
$begingroup$
$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=sumlimits_{(s_1,s_2,s_3)}prod_{i=2}^3binom{s_i+s_{i-1}-1}{s_i}$ since $forall s_1,binom {s_1+s_0-1}{s_1}=binom{s_1}{s_1}=1$. The product just has two terms.
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:02
$begingroup$
@ShubhamJohri, thank you I know it . But, it is not help to me for answer.
$endgroup$
– 1Spectre1
Dec 11 '18 at 6:10
add a comment |
$begingroup$
We know that equation $$s_1+s_2+s_3=n-1 quad mbox{$s_1,s_2,s_3$}geq 1$$
has $binom{n-2}{2}$ solution.
I want to find any good formulae for the following form :
$$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=?$$
where, $s_0=1$ and each $(s_1,s_2,s_3)$ is the solution of above equaiton.
- My all attempts have failed.
$textbf{Edit}:$ I also note that following:
$$n=4Longrightarrow 1$$
$$n=5Longrightarrow 5$$
$$n=6Longrightarrow 18$$
$$n=7Longrightarrow 53$$
$$n=8Longrightarrow 169$$
$$n=9Longrightarrow 502$$
combinatorics discrete-mathematics combinations
asked Dec 10 '18 at 13:28
1Spectre11Spectre1
999
$endgroup$
We know that equation $$s_1+s_2+s_3=n-1 quad mbox{$s_1,s_2,s_3$}geq 1$$
has $binom{n-2}{2}$ solution.
I want to find any good formulae for the following form :
$$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=?$$
where, $s_0=1$ and each $(s_1,s_2,s_3)$ is the solution of above equaiton.
- My all attempts have failed.
$textbf{Edit}:$ I also note that following:
$$n=4Longrightarrow 1$$
$$n=5Longrightarrow 5$$
$$n=6Longrightarrow 18$$
$$n=7Longrightarrow 53$$
$$n=8Longrightarrow 169$$
$$n=9Longrightarrow 502$$
combinatorics discrete-mathematics combinations
combinatorics discrete-mathematics combinations
asked Dec 10 '18 at 13:28
1Spectre11Spectre1
999
asked Dec 10 '18 at 13:28
1Spectre11Spectre1
999
edited Dec 11 '18 at 12:23
1Spectre1
asked Dec 10 '18 at 13:28
1Spectre11Spectre1
999
asked Dec 10 '18 at 13:28
1Spectre11Spectre1
999
asked Dec 10 '18 at 13:28
1Spectre11Spectre1
999
999
3
$begingroup$
$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=sumlimits_{(s_1,s_2,s_3)}prod_{i=2}^3binom{s_i+s_{i-1}-1}{s_i}$ since $forall s_1,binom {s_1+s_0-1}{s_1}=binom{s_1}{s_1}=1$. The product just has two terms.
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:02
$begingroup$
@ShubhamJohri, thank you I know it . But, it is not help to me for answer.
$endgroup$
– 1Spectre1
Dec 11 '18 at 6:10
add a comment |
3
$begingroup$
$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=sumlimits_{(s_1,s_2,s_3)}prod_{i=2}^3binom{s_i+s_{i-1}-1}{s_i}$ since $forall s_1,binom {s_1+s_0-1}{s_1}=binom{s_1}{s_1}=1$. The product just has two terms.
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:02
$begingroup$
@ShubhamJohri, thank you I know it . But, it is not help to me for answer.
$endgroup$
– 1Spectre1
Dec 11 '18 at 6:10
3
3
$begingroup$
$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=sumlimits_{(s_1,s_2,s_3)}prod_{i=2}^3binom{s_i+s_{i-1}-1}{s_i}$ since $forall s_1,binom {s_1+s_0-1}{s_1}=binom{s_1}{s_1}=1$. The product just has two terms.
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:02
$begingroup$
$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=sumlimits_{(s_1,s_2,s_3)}prod_{i=2}^3binom{s_i+s_{i-1}-1}{s_i}$ since $forall s_1,binom {s_1+s_0-1}{s_1}=binom{s_1}{s_1}=1$. The product just has two terms.
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:02
$begingroup$
@ShubhamJohri, thank you I know it . But, it is not help to me for answer.
$endgroup$
– 1Spectre1
Dec 11 '18 at 6:10
$begingroup$
@ShubhamJohri, thank you I know it . But, it is not help to me for answer.
$endgroup$
– 1Spectre1
Dec 11 '18 at 6:10
add a comment |
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$begingroup$
$sumlimits_{(s_1,s_2,s_3)}prod_{i=1}^3binom{s_i+s_{i-1}-1}{s_i}=sumlimits_{(s_1,s_2,s_3)}prod_{i=2}^3binom{s_i+s_{i-1}-1}{s_i}$ since $forall s_1,binom {s_1+s_0-1}{s_1}=binom{s_1}{s_1}=1$. The product just has two terms.
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:02
$begingroup$
@ShubhamJohri, thank you I know it . But, it is not help to me for answer.
$endgroup$
– 1Spectre1
Dec 11 '18 at 6:10