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Source: What is Shared Hosting and what do you need to know!

## Elementary Number Theory – Is it possible to represent a positive integer with a sum of arbitrary many different powers of 3, 5 and 7?

My question is whether it is possible to represent any positive integer as the sum of any number of primes $$3$$. $$5$$, and $$7$$without using a single prime twice. All exponents can not be negative.

(Note that this can be done trivially only with the base $$2$$since this corresponds to writing the number in binary form.)

For example, $$24$$ has two such representations:

$$3 ^ 1 + 3 ^ 2 + 5 ^ 1 + 7 ^ 1 = 3 ^ 0 + 3 ^ 2 + 5 ^ 0 + 5 ^ 1 + 7 ^ 0 + 7 ^ 1 = 24$$

My instinct is that this should not apply, but I could not find a counterexample or a convincing argument against it.

## Presence of a positive measure square in a positive measure set in \$ mathbb {R} ^ 2 \$.

To let $$E subset mathbb {R} ^ 2$$ so that $$overline {E}$$ has a positive effect in $$mathbb {R} ^ 2$$, Can we always find $$A, B subset mathbb {R}$$ so that $$A × B subset E$$ and $$overline {A}, overline {B}$$ is of positive importance in $$mathbb {R}$$?

## Number theory – Euler products in positive form? [reference request]

To let $$K$$ Let's be a global field with a ring of integers $$O$$, and let $$f$$ an integer function whose domain is the set of ideals of $$O$$ (E.g. $$f (I) = | O: I |$$). Extension of the usual definition of the case $$K = mathbb Q$$do you say that $$f$$ is multiplicative if $$newcommand { mfp} { mathfrak {p}} f ( mfp_1 mfp_2) = f ( mfp_1) f ( mfp_2)$$ for any two primes $$mfp_1, mfp_2 triangleleft 1$$, Especially if $$f$$ is then multiplicative and positive $$f (0) = 1$$,

You can define a Dirichlet function associated with it $$f$$:
$$zeta_f (s) = sum_ {I triangleleft O} f (I) ^ {- s} quad (s in C)$$ and assuming that the convergence area is not empty, you get factorization according to Euler:
$$zeta_f (s) = prod_ { mfp ​​in mathrm {Spec} (O) setminus 0} zeta_ {f, mfp} (s)$$
from where $$zeta_ {f, mfp} (s) = sum_ {n ge 0} f ( mfp ^ n) ^ {- ns}$$,

In the case where $$mathrm {char} (K) = 0$$ There is a fairly well-established theory for the analysis of such products, including an interpretation of their convergence abscissa (the infimal real number that limits the range of convergence) in terms of the growth rate of the subtotals
$$sum_ {I triangleleft O, : | O: I |
as well as analytical results that provide estimates for this sum in the case of a certain meromorphic continuation of $$zeta_f$$ exist.

Does anyone know of any parallel results for the case $$mathrm {char} (K)> 0$$? Are there any common references to these types of Euler factorization in positive traits? For example, is the Dedekind zeta function of a function field an examined object, or are there obvious limitations on why one should not try to investigate such a function?

## Calculus – number of positive real roots of \$ a_1x ^ {k_1} (x + 1) ^ {m_1} + a_2x ^ {k_2} (x + 1) ^ {m_2} + ldots + a_nx ^ {k_n} (x + 1) ^ {m_n} -1 \$ is at most \$ n ^ 2 + n + 2 \$

I have the following problem.

Problem. With a positive integer $$n$$, positive integers $$k_1, k_2, ldots, k_n$$. $$m_1, m_2, ldots, m_n$$ and real numbers $$a_1, a_2, ldots, a_n$$, Prove this polynomial
$$a_1x ^ {k_1} (x + 1) ^ {m_1} + a_2x ^ {k_2} (x + 1) ^ {m_2} + ldots + a_nx ^ {k_n} (x + 1) ^ {m_n} -1$$
has at most $$n ^ 2 + n + 2$$ positive real roots.

I know that problems of this kind can usually be solved by applying Descartes' sign rule, rolling theorem or differentiation. For example, the following problem

Problem. With a positive integer $$n$$, positive integers $$k_1, k_2, ldots, k_n$$ and real numbers $$a_1, a_2, ldots, a_n$$, Prove this polynomial
$$a_1x ^ {k_1} + a_2x ^ {k_2} + ldots + a_nx ^ {k_n} -1$$
has at most $$n$$ set real roots.

can be solved by using the character rule of Descartes or by using the set of role with differentiation (just take a derivative and delete the lowest form term) $$x ^ k$$).

However, this approach does not help with the first problem, because after differentiation, the number of terms $$ax ^ k (x + 1) ^ m$$ does not decrease (it can even get twice bigger).

Is there a solution to this problem?

## Agal Algebraic Geometry – Picard Groups of Toric Varieties in Positive Formation

For a toric change $$X _ { Sigma}$$ about the complex numbers assigned to a fan $$Sigma$$ there is a simple short exact sequence that computes the divisor class. To every one-dimensional cone $$rho$$ in the fan is a torusinvarianter disguise $$D _ { rho}$$ (that is the closure of the associated torus run in $$X _ { Sigma}$$). The short exact order is
$$M rightarrow bigoplus _ { rho to Sigma (1)} mathbb {Z} cdot D _ { rho} rightarrow text {Cl} (X _ { Sigma}) rightarrow 0$$ (from where $$M$$ is the sign grid and the sum in the middle is over all the rays that are assigned to the fan.

This description is based on the Orbit Cone equivalent found in Cox Little and Schenck's book (which deals only with toric varieties) $$mathbb {C}$$). The proof they present does not generalize directly to characteristic null fields, and although I have seen that many theorems are applied to algebraically closed fields, I do not realize that this description is generalized. To let $$lambda ^ n$$ denote the Cocharacter associated with $$n$$ in the Cocharacter lattice. For example, the correspondence requires the intersection $$U _ { sigma_1} cap U _ { sigma_2} = U_ { sigma_1 cap sigma_2}$$ and this, in turn, is based on the proposition that $$n$$ lies in a cone $$sigma$$ then and only if $$lim_ {t rightarrow 0} lambda ^ n (t)$$ converges into $$U _ { sigma}$$, This seems to exploit the fact that $$mathbb {C}$$ has a suitable topology.

Is there an example of a smooth fan $$Sigma$$ and a finite field $$mathbb {F}$$ so that the associated toric diversity $$X _ { Sigma, mathbb {F}}$$ has a Picard group that is not isomorphic to the Picard group of $$X _ { Sigma, mathbb {C}}$$or is it true that the Picard group in this case depends only on the fan and not on the definition field? And what about over? $$bar { mathbb {F}}$$?

## Analytic Number Theory – What about a formula that is similar to Mills' formula, but yields positive integers without repeated prime factors?

The wikipedia article for Prime number shows a familiar and curious formula for primes from their section Formula for primesI say the set of mills (see also Wikipedia) mill constant).

Question. I was wondering if you could find or compute a constant and choose an arithmetic function and use a floor or ceiling function to write a formula that generates square-integers, just like Mills formula is a prime-number formula. Many thanks.

So I try to write a formula and unconditionally determine the constant. This formula should produce a square-integer for each $$n geq n_0$$, for certain positive integers $$n_0$$, I would like to know if it is feasible / feasible for a similarly good computational function, see the Exponential of the Mill's formula (I do not know what Wright's sentence looks like, so I'm asking for a formula that Mill & # 39; similar formula, now for integers without repeated prime factors).

I do not know if my question is in the literature. If such a formula exists in the literature, comment on it explicitly with reference to the literature, and I try to search for it and repeat it from the literature.

## nt.number theory – Can we find a true positive sequence \$ (w_ {k}) _ {k≥1} \$ that converges to zero, so that we have an equivalence to the presumption of the Legendres?

The presumption of Legendre states that every positive integer $$k$$ There is at least one prime $$p$$ so that $$k² ,

My question is: Can we find a true positive sequence $$(w_ {k}) _ {k≥1}$$ converge to zero, so we have the following equivalence:

There is a prime number $$p$$ so that $$k² then and only if $$w_ {k} = 0$$,

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Problem: To let $$q$$ Let be a power of a prime. calculation $$| GL (n, q) |$$ in which $$n$$ to be a positive integer.
I did not understand why he came to the conclusion that "There is a total $$q ^ n$$ Line vectors of length $$n$$ over $$F$$Could you explain it to me, thank you all!