Let the set of numbers $A,n$ be of defined as $A_n = \text{the }n\text{'th value of }x\text{ such that }2\not\mid\lfloor{ x^2 - \sqrt{x}}\rfloor$, and $n$ is a positive integer.

So as the first 10 sums are: $$\begin{align} x&&\lfloor{ x^2 - \sqrt{x}}\rfloor\\ 1 &&0\\ 2 &&2\\ 3 &&7\\ 4 &&14\\ 5 &&22\\ 6 &&33\\ 7 &&46\\ 8 &&61\\ 9 &&78\\ 10 &&96\\ \end{align}$$

The terms of $A$ is $A_1 = 3, A_2 =6,A_3 = 8,$ and so on.

I'm trying to determine whether the following is correct:

$$2 n+\lfloor{\frac{1}{2}+\frac{1}{2} \sqrt{8 n - 7}}\rfloor = A_n$$

$A_n = $ Complement of the Connell sequence

This is a real pain, as I have little to no experience in working with the floor function. How can this be (dis)proven?

Any help would be appreciated.

  • $\begingroup$ I think you want to reverse / simplify the definition to the nth positive $x$ for which the floor of $x^2=\sqrt{x}$ is odd. $\endgroup$ – coffeemath Dec 27 '13 at 0:02
  • $\begingroup$ The notation $a|b$ means that $a$ divides $b$, so you should say $2\not |...$. $\endgroup$ – Ragnar Dec 27 '13 at 0:10

I'll just start writing down my thoughts. I don't know where they'll end...
You want $\lfloor{ x^2 - \sqrt{x}}\rfloor$ to be odd. Because you only look at integers $x$, we know that $\lfloor{ x^2 - \sqrt{x}}\rfloor=x^2+\lfloor -\sqrt x\rfloor=x^2-\lceil\sqrt x\rceil$. Now, you want $\lceil \sqrt x\rceil$ to have the opposite parity of $x$ (because $x^2$ has the same parity as $x$. You know that $\lceil\sqrt x\rceil$ is odd when $(2k)^2<x\leq(2k+1)^2$ for some integer $k$. It is even when $(2k+1)^2< x\leq(2k+2)^2$.

Now, we find the following values: $$ 3,6,8,11,13,15,18,20,22,24,27,29,31,33,35,38,\dots $$ As you can see, you get all even or odd values between consecutive squares, and when crossing a square, you change the parity. Also, the differences are always $2$, except when crossing a square; then they are $3$.

Your formula gives these numbers too, so it seems to be correct. (Now, I assume you came this far yourself...)

We know every group of numbers of the same parity is one larger than its predecessor. We want to solve $\frac{k(k+1)}2=n$ to find the number of the group $n$ is in. This is what we want to add to $2n$ to make a correction for the steps of size $3$. Solving it gives $\lfloor\frac{1}{2} \left(\sqrt{8 n+1}-1\right)\rfloor$. We still have to add $1$ to correct for the offset. (this can/should be proven/explained more carefully) Now, we obtain exactly the formula you gave yourself. This is (almost) enough explanation to make it a proof IMO.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.