Orthogonality of Stirling numbers - simplify a variant thereof
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 16 hours ago
Teddy
1,227816
add a comment |
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 16 hours ago
Teddy
1,227816
add a comment |
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 16 hours ago
Teddy
1,227816
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 16 hours ago
Teddy
1,227816
asked 16 hours ago
Teddy
1,227816
asked 16 hours ago
Teddy
1,227816
asked 16 hours ago
Teddy
1,227816
asked 16 hours ago
Teddy
1,227816
1,227816
add a comment |
add a comment |
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