The following sequence has been encrypted:








Can you find the pattern and the next 3 elements?

Hint a - 2nd April 2020

I'm here giving you the next 2 elements. The pattern and the third element remain for your :)



Hint b - 10th April 2020

Factorials are exceedingly used

Hint c - 15th April 2020

I really thought this puzzle will be solved quickly and without hints ^^, it's among the easiest I posted here! Alright, giving you another hint, but it is becoming like... so easy!

At step $n$, there are $n+1$ factorials used!

Hind d - 3rd May 2020

I'm removing the encrypted part of this puzzle with this hint!

The sequence's numbers are:

And so on :)

Hint e - 6th May 2020

Giving you an equivalent function ;)

Defining $H(n)$

This puzzle is about $n+1$ factorials so I told myself, why not use Stirling's formula?
$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$

Suppose that $P(n)$ is the sequence we want to understand. $P$ for Puzzle
$P(1) = 1$
$P(2) = 3$
$P(3) = 60$ and so on.
I'm here going to define $H(n)$, $H$ for Hint, using Stirling's formula and such as $P(n)\sim H(n)$ where the sign $\sim$ means that the two quantities are asymptotic: their ratio tends to 1 as n tends to infinity.

Let $T_n = \dfrac{n(n+1)}2$ and $\displaystyle K(n) = \prod_{k=1}^{n}k^k$ ($K$ is the hyperfactorials) I computed for you: $$H(n) = \dfrac{\pi^{\frac{1-n}2}\sqrt{(n+1)}T_n^{T_n}}{\sqrt{(n-1)!}K(n)}$$

Computing $H(n)$

The following Julia script gave me the first elements of $H(n)$

| P(n) | H(n) rounded at one decimal |
| 1 | 1 |
| 3 | 3.3 |
| 60 | 68.8 |
| 12600 | 14823.8 |
| 37337800 | 45388978.9 |
| 2053230379200 | 2501368479610.3 |
| 24311068187968000 | 3000139818683595001.3 |

function T(n)
$~~~~$ n*(n+1)/2

function K(n)
$~~~~$ k = 1
$~~~~$ for i in 1:n
$~~~~~~~~$ k = k*i^i
$~~~~$ end
$~~~~$ k

function Hint(n)
$~~~~$ pi^((1-n)/2)*sqrt(n+1)*(T(n)^T(n))/(sqrt(factorial(n-1))*K(n)sqrt(2)^n)


And I even made some plots for you :)

enter image description here enter image description here


The fact that the definition of $H(n)$ is really (really) complicated doesn't implies $P(n)$ is also complicated! It's even the opposite, $P(n)$ is simple to express. Think of it like the Stirling's formula, $n!$ is really simple to express and $\sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$ is a little less simple to write.


Please note that I've changed the values of my sequence after Hint d to make it even easier :)

  • $\begingroup$ Are you sure that the sequence is encrypted? It seems like maybe something involving prime numbers but not cryptography... $\endgroup$
    – dan1st
    Mar 27, 2020 at 20:03
  • $\begingroup$ @dan1st, yes, it is. I have encrypted it because, otherwise, it would be too easy finding the pattern :) You made me realize, I forgot the cryptography tag! Many thanks ;) $\endgroup$
    – JKHA
    Mar 27, 2020 at 20:47
  • 2
    $\begingroup$ Heh, I thought your avatar was a piece of the Mandelbrot set until I clicked through to your profile :-) $\endgroup$ Mar 27, 2020 at 20:52
  • $\begingroup$ @Randal'Thor, haha! You did find one of the reason why I chose it :) $\endgroup$
    – JKHA
    Mar 27, 2020 at 20:59
  • $\begingroup$ @Randal'Thor, By the way, I went for yours and the link in your description: riddled.azurewebsites.net isn't working anymore $\endgroup$
    – JKHA
    Mar 28, 2020 at 0:54

1 Answer 1


Hmm. I don't remember seeing this before. Your numbers are


The encryption

simply splits off any final zeros in the decimal representation and gives their count instead of the digits themselves.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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