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

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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). $\endgroup$ – user477343 Feb 14 '18 at 9:41
  • 1
    $\begingroup$ @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 :) $\endgroup$ – user477343 Feb 15 '18 at 9:54
  • 1
    $\begingroup$ 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.


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.