# Generalization of a problem, involving radicals and the floor function, proposed by Ramanujan to the Journal of the Indian Mathematical Society

The section Solved problems from the Wikipedia Floor and ceiling functions shows several problems proposed by Ramanujan ([1]). The purpose of this post, if possible, is try to get the generalization of some of these identities, for positive integers $$n\geq 1$$, involving fractions or radicals and the floor function $$\lfloor x\rfloor$$.

I tried to get a generalization of the identity $$(iii)$$. I don't know if my conjectural identiy is in the literature or has a good mathematical content, these are my previous failed attempts.

Counterexamples for different formulas.

1. A counterexample for the (false) identity $$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor=\lfloor \sqrt[k]{2^k n+k}\rfloor$$ is the integer $$n=525$$ for the case $$k=5$$.

2. Counterexamples for the (false) identity $$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor= \left\lfloor \sqrt[k]{2^k n+2(k-1)} \right\rfloor$$ are the integers $$n=11$$ or $$n=610$$ for the case $$k=6$$.

From this thread of experiments I get the following conjecture.

Conjecture. For each integer $$k\geq 2$$ one has that the identity $$\lfloor \sqrt[k]{n}+\sqrt[k]{n+1}\rfloor=\left\lfloor 2\sqrt[k]{n+\frac{1}{2}}\right\rfloor$$ holds over integers $$n\geq 1$$.

I don't know if it is easy to prove, or if you can to find a counterexample.

Question. Do you know if generalizations (thus with a good mathematical meaning, with mathematical significance) of the mentioned problems proposed by Ramanujan are in the literature? In this case, please refer the literature and I try to find and read from the literature these generalizations of $$(i)$$, $$(ii)$$ or $$(iii)$$. In case that aren't in the literature please add yourself generalization, if possible, with its respective proof for some of those identities. In particular, if you know that my conjeture can be proved or can be refuted by finding a counterexample. Many thanks.

Please if some professor/user finds a counterexample it is welcome that he/she comment it, many thanks. I add for example the following scripts written in Pari/GP as a proof of concept/toy model of my conjecture

for(k=2,10,for(n=1, 1000,if(floor((n)^(1/k)+(n+1)^(1/k))!=floor(2*((n+1/2)^(1/k))),print(k," ",n))))

that you can to evaluate from the web Sage Cell Server, just choose GP as language. Thus aren't showed conunterexamples as outputs. Also you've this one

for(k=2,10,for(n=1, 100,print(floor((n)^(1/k)+(n+1)^(1/k))))) or this

for(k=2,10,for(n=1, 100,print(floor((n)^(1/k)+(n+1)^(1/k))-floor(2*((n+1/2)^(1/k))))))

I add as reference the PARI/GP Developers group of Université Bordeaux 1.

## References:

I believe that the corresponding reference is

[1] Srinivasa Ramanujan, Collected Papers, Question 723 in p. 332, Providence RI: AMS / Chelsea (2000).

[2] I've used also the PARI-GP Reference Card (version 2.2.5), by Karim Belabas (2003), based on an earlier version by Joseph H. Silverman.

• It seems you only need to worry about the cases where $n = \lfloor (r+1/2)^k \rfloor$ for positive integers $r$. Aug 6, 2019 at 17:01
• I hope that the conjecture is true, and the post has good mathematical content. Also I am waiting if some user can propose a generalization of some of those proposed problems by Ramanujan. Many thanks for your contribution @S.Carnahan Aug 6, 2019 at 18:58
• This has somewhat of the flavor of oeis.org/A129935, so a counterexample might be hard to find by brute force. Dec 25, 2020 at 18:10

I can provide a partial proof of your conjecture. It shows that the statement is true for sufficiently large $$n$$, which in this case means that it is true for $$n\geq A_k$$, where $$A_k$$ is a number depending on $$k$$. Exact form of $$A_k$$ will be provided in the proof.

We'll need following ingredients:

1. $$\left\lfloor\sqrt[k]{2^kn+2^{k-1}-1}\right\rfloor =\left\lfloor\sqrt[k]{2^kn+2^{k-1}}\right\rfloor$$

2. $$\sqrt[k]{2^kn+2^{k-1}-1}<\sqrt[k]{n}+\sqrt[k]{n+1}$$, for $$n\geq \left\lceil A_k\right\rceil$$, where $$\left\lceil A_k\right\rceil=\left\{\begin{array}{ll} 0 & \text{for }k=1 \\ 2^{k-3} & \text{for }k\geq2 \end{array}\right.$$

3. $$\sqrt[k]{n}+\sqrt[k]{n+1}<\sqrt[k]{2^kn+2^{k-1}}$$

These three ingredients imply our conjecture. Proofs of individual ingredients provided below.

1. Note that $${2^kn+2^{k-1}-1}$$ and $${2^kn+2^{k-1}}$$ differ only by 1, so the only possibility for 1. not being true is when $$\sqrt[k]{2^kn+2^{k-1}}$$ is an integer. If we assume that $$\sqrt[k]{2^kn+2^{k-1}}=q,\quad\text{for }q\in\mathbb{N},$$ then $$2^kn+2^{k-1}=2^{k-1}(2n+1)=q^k.$$ We can notice, that $$q^k$$ is divisible by $$2^{k-1}$$, but this implies that $$2^k|q^k$$, which can't be true since $$2n+1$$ is odd. Hence 1. is true.

2. Let's rewrite 2. a little bit. We want to prove that $$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}>\sqrt[k]{n+\frac{2^{k-1}-1}{2^k}}.$$ By applying AM-GM inequality we obtain $$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}>\sqrt[2k]{n^2+n}.$$ The inequality $$\sqrt[2k]{n^2+n}\geq\sqrt[k]{n+\frac{2^{k-1}-1}{2^k}}$$ occurs if and only if $$n\geq2n\dfrac{2^{k-1}-1}{2^k}+\left(\dfrac{2^{k-1}-1} {2^k}\right)^2$$, but this inequality implies that $$n\geq \dfrac{2^{2(k-1)}+1-2^k}{2^{k+1}}=:A_k,$$ so if $$n\geq \left\lceil A_k\right\rceil$$, then 2. is true.

3. Let's rewrite 3. in the same manner as we've rewritten 2. We want to prove that $$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}<\sqrt[k]{n+\frac{1}{2}}.$$ By Generalized mean inequality for $$k$$ exponent we obtain $$\frac{\sqrt[k]{n}+\sqrt[k]{n+1}}{2}<\sqrt[k]{\frac{n + n + 1}{2}}=\sqrt[k] {n+\frac{1}{2}}.$$ The equality can't occur, because $$\sqrt[k]{n}\neq \sqrt[k]{n+1}$$ for all $$n\in\mathbb{N}$$. $$\tag*{\blacksquare}$$

The cases, when $$n of course can be checked by computer, but that's not a full proof. From now on I'm working on finding the full proof of it.

• Many thanks, it is incredible. I'm going to study it when I can do it. I would like to dedicate it to your excellence and your colleague (in the top) in comments. Dec 28, 2020 at 18:13
• @user142929 Also, I'm really close to proving another generalization conjecture related to the (ii) Ramanujan Floor problem and I think I might have a better value for $A_k$ in this conjecture. I'll let you know about the progress! Dec 28, 2020 at 19:19
• Many thanks for share your work here (I'm not a professional mathematician), I appreciate it (I add my account in Mathematics Stack Exchange that is user759001 and in Physics Stack Exchange user250478, if you want to read my post in these sites). Best days. Dec 28, 2020 at 20:14
• I've checked manually that the smallest values for $A_k$, starting from $k=1$, are $$(A_k)=(0,0,1,2,3,7,14,28,57,115,233,469,...)$$ which is really close to the one I derived: $$(A_k)=(0,1,1,2,4,8,16,32,64,128,256,512,...)$$ Dec 30, 2020 at 9:08
• oeis.org/A340163 here's a link to the sequence of $A_k$ values. Dec 31, 2020 at 12:36