Orthogonality of Stirling numbers - simplify a variant thereof
$begingroup$
Stirling numbers satisfy the following orthogonality condition. For $k<n$:
$$ sum_{i=k}^n S(n,i) s(i,k)=0,$$
where $s(i,k)$ and $S(n,i)$ are Stirling numbers of the first and second kind, respectively.
My question: Is there a way to simplify the following
$$ sum_{i=k}^n frac{S(n,i) s(i,k)}{2^i}.$$
combinatorics stirling-numbers
asked Jan 4 at 11:55
TeddyTeddy
1,231817
$endgroup$
add a comment |
$begingroup$
Stirling numbers satisfy the following orthogonality condition. For $k<n$:
$$ sum_{i=k}^n S(n,i) s(i,k)=0,$$
where $s(i,k)$ and $S(n,i)$ are Stirling numbers of the first and second kind, respectively.
My question: Is there a way to simplify the following
$$ sum_{i=k}^n frac{S(n,i) s(i,k)}{2^i}.$$
combinatorics stirling-numbers
asked Jan 4 at 11:55
TeddyTeddy
1,231817
$endgroup$
$begingroup$
Is there any reason to believe that this can be simplified?
$endgroup$
– Mike Earnest
Jan 7 at 19:59
add a comment |
$begingroup$
Stirling numbers satisfy the following orthogonality condition. For $k<n$:
$$ sum_{i=k}^n S(n,i) s(i,k)=0,$$
where $s(i,k)$ and $S(n,i)$ are Stirling numbers of the first and second kind, respectively.
My question: Is there a way to simplify the following
$$ sum_{i=k}^n frac{S(n,i) s(i,k)}{2^i}.$$
combinatorics stirling-numbers
asked Jan 4 at 11:55
TeddyTeddy
1,231817
$endgroup$
Stirling numbers satisfy the following orthogonality condition. For $k<n$:
$$ sum_{i=k}^n S(n,i) s(i,k)=0,$$
where $s(i,k)$ and $S(n,i)$ are Stirling numbers of the first and second kind, respectively.
My question: Is there a way to simplify the following
$$ sum_{i=k}^n frac{S(n,i) s(i,k)}{2^i}.$$
combinatorics stirling-numbers
combinatorics stirling-numbers
asked Jan 4 at 11:55
TeddyTeddy
1,231817
asked Jan 4 at 11:55
TeddyTeddy
1,231817
asked Jan 4 at 11:55
TeddyTeddy
1,231817
asked Jan 4 at 11:55
TeddyTeddy
1,231817
asked Jan 4 at 11:55
TeddyTeddy
1,231817
1,231817
$begingroup$
Is there any reason to believe that this can be simplified?
$endgroup$
– Mike Earnest
Jan 7 at 19:59
add a comment |
$begingroup$
Is there any reason to believe that this can be simplified?
$endgroup$
– Mike Earnest
Jan 7 at 19:59
$begingroup$
Is there any reason to believe that this can be simplified?
$endgroup$
– Mike Earnest
Jan 7 at 19:59
$begingroup$
Is there any reason to believe that this can be simplified?
$endgroup$
– Mike Earnest
Jan 7 at 19:59
add a comment |
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$begingroup$
Is there any reason to believe that this can be simplified?
$endgroup$
– Mike Earnest
Jan 7 at 19:59