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Let's assume a limited (and unproved) version of Linnik's theorem: There is a prime $q$ of the form $kp + 1$ for $k \leq (p-2)$ and $p$ a prime. Experimentally this is true, and can be proved for ma …

As I understand the claim $\Lambda(x,n) = \frac{n'}{x'} \Phi(x) \pm V$, it is false for some $n$ and $x$ with $n$ close to $x$. Let us take $x$ to be $P_4=210$, the fourth primorial. Let us take $n$ …

Although this is really a comment, I decided it was important enough to post as an answer. Having studied the problem posted above, I was suspicious of the given claim because of symmetry reasons. It …

Noting that the answers with many components involve few primes, I assume that any answer involves no primes (or denominators with prime factors) greater than 36. The sum of 1/i from m to n can be est …

I used an awk program to generate congruences on n such that, if the Mersenne exponent n satisfied such a congruence, then the corresponding candidate had a small prime factor, which usually was small …

The answers to the MathOverflow question entitled "Optimal Countdown" might be of interest. One poster has results where certain tuples of 6 numbers can yield more than the answers from 1 to 1000. (L …

A faster way would involve looking at the prime factorization of the integer. Let n be pq, where q contains all the 1 mod 4 prime factors of n, and q contains all the 3 mod 4 prime factors. Then the …

I called it $\pi^{-1}(m)$ in a number theory article I posted on the ArXiv. I did not scour the literature, but I conjecture that Erdos never came up with a name for it (he didn't in the papers I saw …

For squarefree $d$, we can translate this into a design problem. Given an $r$ by $c$ array (which correspond to your $r$ many $k$-tuples, but I use $c$ instead of $k$), you need to divide the $k$ dis …

The post above has a link to the term-complexity measure based on size of a term computing a number. The following different model is from my memory of the BCSS paper, so verification would be apprec …

Expanding on the comment, consider the smaller figure where the sum of $x_i$ is restricted to be less than $n$. One gets $\sum_{d \leq s \lt n} {s+1 \choose d-1}$ as a lower bound (or something like …

I assume $p_n \gt m \gt n$ in the inequality. I have a feeling that this will be as challenging as $\pi(x) + \pi(y) \gt \pi(x+y)$ to solve. The essence to me is that composites are sparsest (primes …

Intrigued by the notion in other posts and comments that there might be solutions to this problem involving Hamiltonian paths, I wrote a program to do breadth-first enumeration of such paths for the 3 …

Your sum is just a few terms off from $\sum_{1 \lt j \lt n} (f(j)-f(j-1))(f(n-j)f(n-j-1))$, where hopefully I did not mess up the indices too much. This in turn "looks like it is majorized" by $\int …

Note that x^3 - x = (x-1)x(x+1). Now let x = (a+b+c) and rewrite the equation as (x-1)x(x+1) = 987654320*a + 123456788*b. Let D be gcd(987654320,123456788) = 16. There are integers A, B so that 987 …