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It should be, but I would suggest not in, rather than setminus. In the current presentation, it is hard for me to tell if you mean set difference or quotient. What you call $\langle c\rangle{\cdot}\langle a\rangle{\cdot}\langle c\rangle$ would be more clearly called "Charlie", in my mind. Declaring the alphas to lie outside of Charlie would be more clear to me. Gerhard "Not Doing The Obvious Joke" Paseman, 2015.09.28
Hmm. The average of the d(i) for i from 1 to n is larger than I expected. I will need to check what is counted by log(log(n)). Gerhard "Odds Bigger Than I Thought" Paseman, 2015.09.25
Very small. It is technically d(n) ,the number of divisors of n, divided by n. The quantity d(n) varies from (for n > 1) 2 to log base 2 of n. How d(n) varies over a range has been studied, in general the number is about log(log(n)), and the literature tells you how it varies from this. Sometimes $\tau$ is used instead of d for this function. Gerhard "Check Out The Chances Online" Paseman, 2015.09.25
Thank you for addressing some of the mathematics. Can you recommend something that would encourage students to read the paper, or that you took away from the paper? ("No" would be a reasonable answer.) Gerhard "Looking For The Bright Side" Paseman, 2015.09.25
I concur with Felipe. Much as I am interested in the topic, there are cases that can be used to test such conjectures. In particular, the Lehmer paper and Aaron Meyerowitz's question (88777) and answer show a worst case scenario: for a product x of k primes of the form (mt - 1) a y near x/m produces an error of near 2^{k-1}. You can do this testing and some additional research on your own for a while. Leave editing the question alone for a week or more until you have a significant idea to add/take away. Gerhard "Resist The Temptation Of Tweaking" Paseman, 2015.09.25
This needs a lot of cleanup. You need to resolve occurrences of n,x, and y. My guess is that wherever you have n you mean one of x or y instead. Also, you need to say lambda(x,y) means count of totatives of x in interval (0,y], or whatever it is you mean. Finally, test it by choosing y so that x_sharp is small, and see if you can break your claim. Gerhard "So Practice What I Preach" Paseman, 2015.09.24
Alternatively, one could use a slowly branching tree. Let (1) be the root, and for any member s with terminal element n, throw in s concat (2n) and s concat (2n+1). One can tweak the branching rate and other levels to push c to 1/2 or perhaps below. Gerhard "Construction Lovely As A Tree" Paseman, 2015.09.23
If you choose x to be a larger primorial, you can get n (say n+6=x) not far from x to be squarefree, and you will run into the same problem again. I suggest you look through your proof and use these choices of x and n to find where the proof breaks. Gerhard "Always, Always Use Test Cases" Paseman, 2015.09.23
I made a mistake: $\frac{x'-n'}{x'n'}$ differs from the result I use by a factor of 2 for my choices of $x$ and $n$. However, the result when corrected still shows no integer in the range, and the claim is still false. Gerhard "Correct Up To Small Error" Paseman, 2015.09.23
There is the generic situation: if Omega is all r-sets of an n-set $\binom{n-r}{r}$ do not intersect a given r-set $A$, giving a ratio of $(a-b)/a$ where $a=n(n-1)...(n-r+1)$ and $b=(n-r)(n-r-1)...(n-2r+1)$. In general $f_{r,c}(n)$ won't have much range for large $n$ and small $r$. Gerhard "How Tight Do You Need?" Paseman, 2015.09.22
