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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 …

Although I am also interested in the number of distinct prime factors (not counting multiplicity), today I use $\omega(m)$ to denote the number of (positive) prime factors (with multiplicity) of the i …

Since you include 0 and do not ask that the a_i are increasing with i, your questions are morally equivalent to counting lattice points in the m dimensional plane $n= \sum x_i,$ except you want those …

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 …

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 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 $ …

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 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 …

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 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 …

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 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 …

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, …

In response to a comment posted under Powers of $2$ and the products of initial odd primes , I shall raise some questions about quantities near $O_n= P_{n+1}/2$, the product of the first $n$ odd prim …

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 …