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I was playing around with numbers and I discovered that $$\left\lfloor e^{\frac{p_4\#}{p_5}}\right\rfloor=\left\lfloor e^{\frac{210}{11}}\right\rfloor =13981^2,$$ with floor function $\lfloor x \rfloor := \max\{m\in\mathbb{Z} : m\leqslant x\}$; $ \ p_n$ denotes the $n^{\text{th}}$ prime number; $ \ $and primorial $$p_n\#_c := \prod_{i=1}^n (p_i + c).\tag{$p_n\#_0 = p_n\#$}$$

Furthermore, $$\left\lfloor e^{\frac{p_1\#}{p_2}}\right\rfloor = 1^2.$$ I then made a conjecture with very few support, that $$\left\lfloor e^{\frac{p_{n^2}\#}{p_{n^2 + 1}}}\right\rfloor \tag1$$ is always a square number, say $k_n^{\ \ 2}$.

Does somebody have a big enough computer to find the value of $(k_n)_{n\geqslant3}$? Or can my conjecture be proven/disproven with a pen and paper? And perhaps, to support the potential of this not being a coincidence, every divisor of $13981$ takes the form $(13981 - 10x)$ for some $x\geqslant 0$. Maybe if this conjecture is true, $k_n > 1$ has this certain property?

Edit 1: With some computational power, I found that $k_3^{\ \ 2}$ has $\approx 176,000$ digits. It turned out, however, that I was wrong, and really, $k_3^{ \ \ 2}$ has $\approx 3,340,970$ digits.

Edit 2: Expressions equivalent to $(1)$, avoiding $e$, for any base $b\in\mathbb{N}_{>1}$:$$\left\lfloor b^{\frac{p_{n^2}\#}{p_{n^2 + 1} \cdot Ln(b)}}\right\rfloor=k_n^{ \ \ 2}$$E.g. base $b=2$:$$\left\lfloor 2^{\frac{p_{n^2}\#}{p_{n^2 + 1} \cdot Ln(2)}}\right\rfloor=k_n^{ \ \ 2}$$E.g. base $b=n^2$:$$\left\lfloor (n^2)^{\frac{p_{n^2}\#}{p_{n^2 + 1} \cdot Ln(n^2)}}\right\rfloor=k_n^{ \ \ 2}$$

Mr Pie
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  • My computer said it was equal to some decimal $\times , 10^{17580}$. Is it possible to tell, though, whether or not that number is a square? – Mr Pie Feb 14 '18 at 08:58
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    You didn't perchance use your computer's calculator? For huge numbers, you need software like Mathematica (which I use). There are probably clever number-theoretic tests to tell whether a huge number is a square, and I faintly remember a discussion like that here in MSE. – Tito Piezas III Feb 14 '18 at 09:02
  • @TitoPiezasIII Yes, I did use a huge calculator that I downloaded because that was all I could find. It took $9$ hours for it to come up with that result... At least I know that every square number has to end with a digit $0, 1, 4, 5, 6$ or $9$, and the digital root of a square is always $1, 4, 7$ or $9$. – Mr Pie Feb 14 '18 at 09:04
  • My apologies, I meant, $$\left\lfloor\exp\bigg(\frac{p_9#}{p_{10}}\bigg)\right\rfloor=\left\lfloor\exp\bigg(\frac{223092870}{29}\bigg)\right\rfloor \approx 3.97884\times10^{3340965}$$ You can verify that result in Wolfram which computes it in seconds – Tito Piezas III Feb 14 '18 at 09:10
  • @TitoPiezasIII Oh my, that is even bigger... – Mr Pie Feb 14 '18 at 09:15
  • The first value I gave was just its square root. – Tito Piezas III Feb 14 '18 at 09:16
  • @TitoPiezasIII wait really? That means it is a square. If it wasn't, it would be irrational, but it is not because it was a decimal multiplied by $10^n$ which means it is finite. Right? Oh wait.... it is also approximate... – Mr Pie Feb 14 '18 at 09:17
  • No, no. It was an approximate square root. Sorry. :) – Tito Piezas III Feb 14 '18 at 09:18
  • @TitoPiezasIII hahah right when I edit my comment to mention that, you come in... Thank you though. At least you cleared up an error :) – Mr Pie Feb 14 '18 at 09:19
  • @TitoPiezasIII I think when you were referring to a discussion about large numbers being square numbers, you might have referred to this link $\longrightarrow$ https://math.stackexchange.com/questions/41337/efficient-way-to-determine-if-a-number-is-perfect-square – Mr Pie Feb 14 '18 at 09:23
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    Yes, that's it. I just realized it considered 600 digits as "large". Yours has more than 3 million digits, so you may have a slight problem applying the algorithm. :) – Tito Piezas III Feb 14 '18 at 09:27
  • @user477343 quite interesting... I would try to focus it as a "square generating function" in the fashion of Mills' constant and its prime generating function. In this case $e$ is the constant and the intervals that make possible the sequential embedding of squares would be related with the relationship between the primorial and the next prime... I will try to check it if I have time this week, sounds interesting (if no counterexamples are found!) https://en.wikipedia.org/wiki/Mills%27_constant – iadvd Feb 14 '18 at 09:28
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    @iadvd and it is funny because I was thinking to myself before I discovered something like this, what if I just played around with numbers? and then all of a sudden, bloop there you go. I have also heard of Mills' constant, and have tried making approximations of it (because a formula for it has not been discovered, as far as I know). – Mr Pie Feb 14 '18 at 09:41
  • @user477343 I have added to your question some equivalent expressions trying to avoid the use of $e$ as the specific base. For instance base $2$ and base $n^2$ might be also interesting to understand the behavior. – iadvd Feb 15 '18 at 04:36
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    @iadvd I like the $\exp(\cdot)$ better but I'm sure everyone gets the idea, and that is the whole point anyway, so i'm not too fussed :) – Mr Pie Feb 15 '18 at 09:54
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    ok I think I might be able to have crack at this one – Adam Ledger Oct 26 '18 at 00:01
  • @Adam thanks! Please try... this has been bugging me for a while... I encourage partial answers, at least :) – Mr Pie Oct 27 '18 at 08:00
  • @user477343 sure we that's generally how I work with my own questions, I've posted quite a few now that have partial or near complete answers, unfortunately with number theory conjectures that are expressible with first order logic you just can't claim QED unless the proof is direct – Adam Ledger Oct 27 '18 at 08:27
  • OP, is there any reason you would expect this conjecture to be true? This isn't anything close to a proof/disproof but with just 2 examples it hardly seems like something one would expect to be true! – YiFan Tey Oct 27 '18 at 13:49

1 Answers1

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Conjectured "formulas" such as this are most likely the result of coincidence as one can easily find a constant $A$ and a function $f$ such that the values of

$$ g(n)=\lfloor A^{f(n)} \rfloor $$

are within a given set $S=\{s_0, s_1, \ldots\}$, by knowing the density of $S$. For instance, in the case of Mills' prime-generating formula, it is known that there is always a prime in the interval $[(P-1)^3,P^3]$ for $P>1$, so there must exist a positive nonzero constant $A$ such that

$$ m(n)=\lfloor A^{3^n}\rfloor $$

yields only primes.

Moreover, don't forget that seemingly remarkable mathematical coincidences are easy to generate, so by "playing around" one might find formulas such as yours without any real meaning.

Klangen
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