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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions

-1 votes

Upper bound for number of prime numbers in a range

The worst case is if the primes are spread out, so that there is at least (log x)/2 between each pair of primes (and log log x > 2), so the best case (if you admit reasoning by analogy) is if the prim …
The Masked Avenger's user avatar
2 votes

Splitting integers 1, 2, 3, … n to avoid least possible sum

As n gets larger, the quantity g(n) - 2n should grow, primarily because small sums can't be avoided by an even partition. An analysis that g(6) > 13 should illustrate the principle. To attempt to av …
The Masked Avenger's user avatar
2 votes

Are these inequalities for primes equivalent?

I am using Anthony Quas's reformulation to restate the problem. Letting the $n$th prime gap be given by $d_n= p_{n+1} - p_n$, and given $n$ we relabel $a=d_n$ and $b=d_{n+1}$, we look at $L$ as the s …
1 vote

A limit concerning prime numbers

A theorem of Zsigmondy says that for $n$ large enough (which means $n$ at least 7 in the worst case, for $r=2$), $r^n - 1$ has a prime factor $q$ which does not divide $r^m - 1$, for any $m$ less than …
The Masked Avenger's user avatar
1 vote

Primes which divide exactly one odd composite in a sequence of consecutive odd composites

Let $k$ be the number of odd composites in the interval. I temporarily extend the notion of composite to include nonprimes. Suppose the odd numbers have no prime factors. Then they are 1 and -1, th …
2 votes
Accepted

Products of relative prime numbers with least sum

Notice that the product prod P ( bounded below by n) represents the order of a permutation with cycle structure given by P and sitting in S_m, where m=f(n). So considering the largest order of elemen …
The Masked Avenger's user avatar
0 votes

A prime sequence can be partitioned into two sets of equal or consecutive sum

Expanding on the comment above, consider Pn, the set of the first n primes, and SSn, the set of subset sums of Pn. For n greater than 3, we see that SSn is 6 numbers shy of being the interval [0, Sn], …
The Masked Avenger's user avatar
4 votes

The diameter of a certain graph on the positive integers

Maybe this will work. Given positive integers a and b, choose c large enough so that c^2 > a+b. also, choose c so that c^2 -a -b is odd and factors as (e+d)(e-d). Then a has an edge with c^2 - a, b h …
The Masked Avenger's user avatar
1 vote

Is there a lower bound for the first non-trivial sequence of consecutive integers where each...

Your conditions seem to imply a search for a confluence of a sizable prime gap in which a not very smooth number (one with least prime factor of $p_n$) occurs. You can limit the search by looking "be …
The Masked Avenger's user avatar
2 votes
Accepted

Arithmetic progression and most significant digits in different bases

The problem is a little harder than it seems at first glance. Pick m large, say m > 8. There are 3k=3^m numbers with first ternary digit 1 and m other ternary digits. Suppose 0<= a < 3^m is smallest s …
The Masked Avenger's user avatar
0 votes

Lower bound for Euler's totient for almost all integers

Given $n$, the set of integers $m$ coprime to $n$ has nonzero asymptotic density, thus so does the set $mn$. But then $\phi(n)/n > \phi(mn)/mn$ , so the $\liminf_n\phi(n)/n$ will remain the same off …
The Masked Avenger's user avatar
2 votes

Conjecture on the square root of the sum of the squares of the prime factors of a number

It may be of interest to consider in general when A=A_n is integral. I will assume n is given and drop the subscript. A is integral when n =p^k, for p prime and k a square. A integral and n composi …
The Masked Avenger's user avatar
4 votes

distribution of coprime integers

In a similar question Bound the error in estimating a relative totient function , Alan Haynes notes in an answer that Vijayraghavan in 1951 had published a result showing many $n$ for which (for certa …
The Masked Avenger's user avatar
16 votes
Accepted

Arbitrarily long arithmetic progressions

A simple proof is available as well. Pick p coprime to d and let t be such that td=1 mod p. Then, mod p, t times the arithmetic progression looks like a sequence of consecutive integers. Thus its l …
The Masked Avenger's user avatar
1 vote

Covering a finite subset of $\mathbb{N}$ with prime arithmetic progressions

This is equivalent to studying large intervals of numbers that are not coprime to a product of some selected prime numbers. When one has such a large interval, one can translate it by subtraction to …
The Masked Avenger's user avatar

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