Or ( converting string operations to arithmetic ) if t is such a number, so is (t - 27)/1000 . (Unless I need to remove the most significant digits instead.)

Please change the title to include the question. Even "On a difference of differences of consecutive primes" is not specific enough.

Near duplicate: mathoverflow.net/questions/137287/… .

@Mirko, interesting. I will look at your posted followup.

(6k +- 1)^2 is 1 mod 24, so there will be at least four halvings of p^2 for prime p greater than 5. It might be worthwhile to look at the dynamics mod 210 or mod 2310.

You have 3 versions leading to 3 possible conjectures. I think my answer says all 3 conjectures are false. As your conjectures deal with how sparse certain multiples of x are in unions, they seem to speak to how many multiples should appear outside a union, and thus how few primes there are in a given interval. I think this goes against recent results, probably on small prime gaps, and that to form a good conjecture, you need to think about these results. You may also need to put more constraints on the sets A_i in order to get a conjecture that is harder to refute.

The counterexample in the answer also addresses the last version. I think what you will find is that your hope runs counter to recent work on large gaps between primes. That, or you need more conditions on the A's.

As an example to tweak, pick d_i to be primes that are 3 mod 7, and pick enough of them so that D is 1 mod 7. Then D +2d_i is a multiple of 7, and one can go from 1/3 density by choosing just 3 elememts for each A_i, to almost 1/2 density for the union. And this isn't even exploring situations where the intersection of A_i and A_j has more than 1 element.

I note that this applies to both versions of the conjecture. If the goal is to bound from above the density of multiples of x in a union, one has to look at how the summands share elements so as not to skew the density results, which is the point of the answer above.

@XuexingLu you need to leave off the space after @ for notification to work. Also, my powers of 3 example refutes your conclusion for d =7/2 and satisfies conditions 1, 3 and 4. I suspect one can show that your conclusion does not follow from 1,2,3 and 4.

More specifically, take Ai to be 3^i times 1,2,3,4. The union will have almost every third entry be a multiple of 4.

The problem is that some terms can appear in each A_i. Try powers of 3. One can have the union end with 27 36 54 81 108, which raises the frequency of multiples of 4 a bit over 1/4.

In fact, here is a start at a cooperative play strategy: Match up the 8(k-1) points first, then arrange for a special matching among the remaining. In actual competitive play of the game, I imagine it is easy to knock others or yourself off, so that all 8 players surviving has a likelihood much smaller than 1 over (8k choose 8).

Just to make sure I got it right, h() returns the supremum ( when such a supremum exists) of the y-coordinate values of its input?

Using a hyperplane, bisect a general point set into the n points for your knot, and the rest. If your unknot argument holds, build an unknot and the desired knot, and then tie the two together.

Further, if one is asking $\Omega(n)$ queries, one can use $n/2$ queries to determine the identity element if its "identity" is not known. (Pun intended.)