Stirling number congruence $Sleft(n, frac{p-1}{2}right) pmod{p}$
For $p$ prime, is there a simple expansion for the Stirling number of the second kind $S(n,k)$, of the form $Sleft(n,frac{p-1}{2}right) pmod{p}$? I tried using the explicit expansion $$left( frac{p-1}{2} right)!S(n,frac{p-1}{2}) = (-1)^{frac{p-1}{2}}sum_{l=0}^{frac{p-1}{2}}(-1)^l binom{frac{p-1}{2} }{l}l^n$$ and a literature search, but got nowhere.
For context: the von Staudt - Clausen theorem exactly describes the fractional part of the $n$-th Bernoulli number $B_n$, by writing $B_n$ as a sum over Stirling numbers and reducing mod $p$. This means that we have to analyze $S(n,p) pmod{p}$, which is either $0$ or $1$. I'm looking at a specialization of poly-Bernoulli numbers, which are multiple sums over Bernoulli numbers that arise when we try to generalize the formula for the Riemann zeta function $zeta(2n) = (-1)^{n+1} frac{B_{2n}(2pi)^{2n}}{2(2n)!}$. Finding an analog of the von Staudt - Clausen theorem ends up reducing to analyzing the Stirling numbers $Sleft(n,frac{p-1}{2}right) pmod{p}$, and an explicit expression would give us some arithmetic information about a special kind of multiple zeta function.
Numerically, it looks like $Sleft(n,frac{p-1}{2}right) pmod{p}$ is periodic with period $p-1$, but I still can't conjecture the actual value.
number-theory modular-arithmetic stirling-numbers
asked Nov 29 at 6:38
twakhare
505
add a comment |
For $p$ prime, is there a simple expansion for the Stirling number of the second kind $S(n,k)$, of the form $Sleft(n,frac{p-1}{2}right) pmod{p}$? I tried using the explicit expansion $$left( frac{p-1}{2} right)!S(n,frac{p-1}{2}) = (-1)^{frac{p-1}{2}}sum_{l=0}^{frac{p-1}{2}}(-1)^l binom{frac{p-1}{2} }{l}l^n$$ and a literature search, but got nowhere.
For context: the von Staudt - Clausen theorem exactly describes the fractional part of the $n$-th Bernoulli number $B_n$, by writing $B_n$ as a sum over Stirling numbers and reducing mod $p$. This means that we have to analyze $S(n,p) pmod{p}$, which is either $0$ or $1$. I'm looking at a specialization of poly-Bernoulli numbers, which are multiple sums over Bernoulli numbers that arise when we try to generalize the formula for the Riemann zeta function $zeta(2n) = (-1)^{n+1} frac{B_{2n}(2pi)^{2n}}{2(2n)!}$. Finding an analog of the von Staudt - Clausen theorem ends up reducing to analyzing the Stirling numbers $Sleft(n,frac{p-1}{2}right) pmod{p}$, and an explicit expression would give us some arithmetic information about a special kind of multiple zeta function.
Numerically, it looks like $Sleft(n,frac{p-1}{2}right) pmod{p}$ is periodic with period $p-1$, but I still can't conjecture the actual value.
number-theory modular-arithmetic stirling-numbers
asked Nov 29 at 6:38
twakhare
505
1
An explicit expansion should read $Sleft(n,frac{p-1}{2}right)=frac{(-1)^frac{p-1}{2}}{left(frac{p-1}{2}right)!}sum_{ellge 0}(-1)^ellbinom{frac{p-1}{2}}{ell}ell^n$, right?
– René Gy
Nov 30 at 19:49
my bad -- fixed it!
– twakhare
Dec 3 at 1:07
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
– Carl Mummert
Dec 6 at 17:49
add a comment |
For $p$ prime, is there a simple expansion for the Stirling number of the second kind $S(n,k)$, of the form $Sleft(n,frac{p-1}{2}right) pmod{p}$? I tried using the explicit expansion $$left( frac{p-1}{2} right)!S(n,frac{p-1}{2}) = (-1)^{frac{p-1}{2}}sum_{l=0}^{frac{p-1}{2}}(-1)^l binom{frac{p-1}{2} }{l}l^n$$ and a literature search, but got nowhere.
For context: the von Staudt - Clausen theorem exactly describes the fractional part of the $n$-th Bernoulli number $B_n$, by writing $B_n$ as a sum over Stirling numbers and reducing mod $p$. This means that we have to analyze $S(n,p) pmod{p}$, which is either $0$ or $1$. I'm looking at a specialization of poly-Bernoulli numbers, which are multiple sums over Bernoulli numbers that arise when we try to generalize the formula for the Riemann zeta function $zeta(2n) = (-1)^{n+1} frac{B_{2n}(2pi)^{2n}}{2(2n)!}$. Finding an analog of the von Staudt - Clausen theorem ends up reducing to analyzing the Stirling numbers $Sleft(n,frac{p-1}{2}right) pmod{p}$, and an explicit expression would give us some arithmetic information about a special kind of multiple zeta function.
Numerically, it looks like $Sleft(n,frac{p-1}{2}right) pmod{p}$ is periodic with period $p-1$, but I still can't conjecture the actual value.
number-theory modular-arithmetic stirling-numbers
asked Nov 29 at 6:38
twakhare
505
For $p$ prime, is there a simple expansion for the Stirling number of the second kind $S(n,k)$, of the form $Sleft(n,frac{p-1}{2}right) pmod{p}$? I tried using the explicit expansion $$left( frac{p-1}{2} right)!S(n,frac{p-1}{2}) = (-1)^{frac{p-1}{2}}sum_{l=0}^{frac{p-1}{2}}(-1)^l binom{frac{p-1}{2} }{l}l^n$$ and a literature search, but got nowhere.
For context: the von Staudt - Clausen theorem exactly describes the fractional part of the $n$-th Bernoulli number $B_n$, by writing $B_n$ as a sum over Stirling numbers and reducing mod $p$. This means that we have to analyze $S(n,p) pmod{p}$, which is either $0$ or $1$. I'm looking at a specialization of poly-Bernoulli numbers, which are multiple sums over Bernoulli numbers that arise when we try to generalize the formula for the Riemann zeta function $zeta(2n) = (-1)^{n+1} frac{B_{2n}(2pi)^{2n}}{2(2n)!}$. Finding an analog of the von Staudt - Clausen theorem ends up reducing to analyzing the Stirling numbers $Sleft(n,frac{p-1}{2}right) pmod{p}$, and an explicit expression would give us some arithmetic information about a special kind of multiple zeta function.
Numerically, it looks like $Sleft(n,frac{p-1}{2}right) pmod{p}$ is periodic with period $p-1$, but I still can't conjecture the actual value.
number-theory modular-arithmetic stirling-numbers
number-theory modular-arithmetic stirling-numbers
asked Nov 29 at 6:38
twakhare
505
asked Nov 29 at 6:38
twakhare
505
edited Dec 7 at 3:18
asked Nov 29 at 6:38
twakhare
505
asked Nov 29 at 6:38
twakhare
505
asked Nov 29 at 6:38
twakhare
505
505
1
An explicit expansion should read $Sleft(n,frac{p-1}{2}right)=frac{(-1)^frac{p-1}{2}}{left(frac{p-1}{2}right)!}sum_{ellge 0}(-1)^ellbinom{frac{p-1}{2}}{ell}ell^n$, right?
– René Gy
Nov 30 at 19:49
my bad -- fixed it!
– twakhare
Dec 3 at 1:07
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
– Carl Mummert
Dec 6 at 17:49
add a comment |
1
An explicit expansion should read $Sleft(n,frac{p-1}{2}right)=frac{(-1)^frac{p-1}{2}}{left(frac{p-1}{2}right)!}sum_{ellge 0}(-1)^ellbinom{frac{p-1}{2}}{ell}ell^n$, right?
– René Gy
Nov 30 at 19:49
my bad -- fixed it!
– twakhare
Dec 3 at 1:07
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
– Carl Mummert
Dec 6 at 17:49
1
1
An explicit expansion should read $Sleft(n,frac{p-1}{2}right)=frac{(-1)^frac{p-1}{2}}{left(frac{p-1}{2}right)!}sum_{ellge 0}(-1)^ellbinom{frac{p-1}{2}}{ell}ell^n$, right?
– René Gy
Nov 30 at 19:49
An explicit expansion should read $Sleft(n,frac{p-1}{2}right)=frac{(-1)^frac{p-1}{2}}{left(frac{p-1}{2}right)!}sum_{ellge 0}(-1)^ellbinom{frac{p-1}{2}}{ell}ell^n$, right?
– René Gy
Nov 30 at 19:49
my bad -- fixed it!
– twakhare
Dec 3 at 1:07
my bad -- fixed it!
– twakhare
Dec 3 at 1:07
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
– Carl Mummert
Dec 6 at 17:49
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
– Carl Mummert
Dec 6 at 17:49
add a comment |
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1
An explicit expansion should read $Sleft(n,frac{p-1}{2}right)=frac{(-1)^frac{p-1}{2}}{left(frac{p-1}{2}right)!}sum_{ellge 0}(-1)^ellbinom{frac{p-1}{2}}{ell}ell^n$, right?
– René Gy
Nov 30 at 19:49
my bad -- fixed it!
– twakhare
Dec 3 at 1:07
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
– Carl Mummert
Dec 6 at 17:49