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 '20 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 '20 at 20:47
  • 2
    $\begingroup$ Heh, I thought your avatar was a piece of the Mandelbrot set until I clicked through to your profile :-) $\endgroup$ – Rand al'Thor Mar 27 '20 at 20:52
  • $\begingroup$ @Randal'Thor, haha! You did find one of the reason why I chose it :) $\endgroup$ – JKHA Mar 27 '20 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 '20 at 0:54

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.


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