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Nov 25 |
answered | short character sums averaged on the character |
Nov 25 |
reviewed | Leave Open Underlying idea for (automorphic) L-function? |
Nov 25 |
comment |
lower-bound for $Pr[X\geq EX]$
So I had in mind the situation that the variables are all independent. It would be good for OP to clarify if that's to be assumed or not. The situation when the variables are independent is related to problems in combinatorial number theory (studied by Alladi, Erdos and Vaaler), and to the Manickam, Miklosh and Singhi conjecture in combinatorics (on which there has been interesting progress lately). |
Nov 25 |
reviewed | Leave Open Determinant of matrix from set {-1, 1} |
Nov 24 |
comment |
lower-bound for $Pr[X\geq EX]$
This is actually a non-trivial and very interesting question. It should not be closed. I'll try to add an answer with references to related work if I get a chance. |
Nov 24 |
reviewed | Leave Open lower-bound for $Pr[X\geq EX]$ |
Nov 24 |
reviewed | Close Weighted Distribution |
Nov 23 |
comment |
Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
Seva: It doesn't seem to me that $p$ is assumed to be $1\pmod 8$. When $q$ divides $a$ (which is the example you raised) then the biquadratic symbol is determined by $(\frac{2}{q})$. I'm not an exper, and it may be safest to consult Lemmermeyer's book. |
Nov 23 |
comment |
Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
The sign doesn't matter. If you multiply $(\frac{\sigma(b+\sigma)}{q})$ and $(\frac{-\sigma(b-\sigma)}{q})$ together, you get $(\frac{-\sigma^2(b-\sigma^2)}{q}) = (\frac{a^2}{q})=1$. |
Nov 23 |
comment |
Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
An answer is given in the wikipedia page you linked in the section under Dirichlet. If $p\equiv \sigma^2 \pmod q$ then $(\frac{-q}{p})_4=(\frac{\sigma(b+\sigma)}{q})$. A reference is given to Lemmermeyer's book. |
Nov 22 |
reviewed | Leave Closed Robotics, Cryptography, and Genetics applications of Grothendieck's work? |
Nov 22 |
answered | A perfect $(n,k)$ shuffle function |
Nov 22 |
comment |
A perfect $(n,k)$ shuffle function
If you label the cards $0$ to $n-1$ then your shuffle corresponds to multiplying card $i$ by the inverse of $k$ modulo $n-1$. Thus it returns to the original configuration in the order of $k^{-1}$ mod $n-1$ steps. But this is the same as the order of $k$ mod $n-1$. |
Nov 22 |
comment |
A perfect $(n,k)$ shuffle function
Seems to be the order of $k$ mod $n-1$. (This is well known for the usual perfect outer shuffle, and seems numerically to work in your other examples.) |
Nov 20 |
reviewed | Close “Almost” zeta function |
Nov 20 |
reviewed | No Action Needed Normal Covering of a Finite Group |
Nov 19 |
reviewed | Leave Open Finite extension of fields with no primitive element |
Nov 19 |
reviewed | Close Robotics, Cryptography, and Genetics applications of Grothendieck's work? |
Nov 17 |
reviewed | Leave Closed Is there a unique solution? |
Nov 16 |
reviewed | Close What does analyticity imply in complex analysis? |