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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?