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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 question makes sense if the $r$'s are all positive. The answer is likely to be yes for the folllowing reason: if it were no, there would be an explicit sequence where the $r$'s grow faster than $ …

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 …

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 …

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 …

The result is wrong. The major problem is that there is little control over the common elements of the sets A_i. I suggest coming up with simpler sets of conditions and checking relations between th …

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 …

Does the following approach (or something near it) exist in the number theory literature? I will provide some motivation for $\omega(p^n - 1)$ as $n \rightarrow \infty$ and for this question. First, …

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 …

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], …

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 …

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 short answer is no. It is likely to be sufficient, although right now the best known is actually that pi(x + x^{0.525}) > pi(x) for all large x (and likely all x > 117). If it were necessary, thi …

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 more generalized problem (given window positions, and alphabet size, find a cyclic or even non cyclic zequence such that shifting window positions gives all words of length same as concatenated wi …