Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 50509

Questions in which polynomials (single or several variables) play a key role. It is typically important that this tag is combined with other tags; polynomials appear in very different contexts. Please, use at least one of the top-level tags, such as nt.number-theory, co.combinatorics, ac.commutative-algebra, in addition to it. Also, note the more specific tags for some special types of polynomials, e.g., orthogonal-polynomials, symmetric-polynomials.

8 votes
3 answers
376 views

Self-reciprocal polynomials over finite fields

Let $SRMI_q(2n)$ denote the number of self-reciprocal irreducible monic polynomials of even degree $2n$ over the finite field $\mathbf{F}_q$ with $q$ elements. … have in mind is an analogue of the well-known formula \begin{equation*} \prod_{n \geq 1} \frac{1}{(1-x^n)^{I_q(n)}} = \frac{1}{1-qx} \end{equation*} where $I_q(n)$ is the number of irreducible monic polynomials
Jesper M. Moller's user avatar
3 votes

Transforming numbers of irreducible polynomials

Here is a reformulation of Ofir Gorodetsky's excellent answer to my question: Let $(a(n))_{n \geq 1}$ be a sequence of rational numbers. Define the transformed sequence $T(a)$ of $a$ to have $n$th ele …
Jesper M. Moller's user avatar
4 votes
2 answers
612 views

Transforming numbers of irreducible polynomials

Let $M(n)(q)$, where $q$ is a prime power and $n$ a natural number, stand for the number of irreducible monic polynomials of degree $n$ in the polynomial ring $\mathbf{F}_{q}[X]$ over the finite field … Consider the sum of polynomials in $\mathbf{Q}[q]$ \begin{equation*} \sum_{[n_1^{e_1},\ldots, n_s^{e_s}] \in P(n)} (-1)^{\sum e_i} \prod_{i=1}^s \binom{M(n_i)}{e_i} \end{equation*} ranging over all …
Jesper M. Moller's user avatar