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 $b=n^2$:$$\left\lfloor (n^2)^{\frac{p_{n^2}\#}{p_{n^2 + 1} \cdot Ln(n^2)}}\right\rfloor=k_n^{ \ \ 2}$$

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    $\begingroup$ 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. $\endgroup$ – Tito Piezas III Feb 14 '18 at 9:02
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    $\begingroup$ 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. :) $\endgroup$ – Tito Piezas III Feb 14 '18 at 9:27
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    $\begingroup$ @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). $\endgroup$ – Feeds Feb 14 '18 at 9:41
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    $\begingroup$ @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 :) $\endgroup$ – Feeds Feb 15 '18 at 9:54
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    $\begingroup$ ok I think I might be able to have crack at this one $\endgroup$ – Adam Oct 26 '18 at 0:01

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.


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