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You are watching: 1-2-6-24-120

I was playing v No Man"s Sky once I ran into a series of numbers and was asked what the following number would be.

$$1, 2, 6, 24, 120$$

This is for a terminal assess password in the game no man sky. The 3 options they provide are; 720, 620, 180  The next number is $840$. The $n$th hatchet in the sequence is the smallest number v $2^n$ divisors.

Er ... The next number is $6$. The $n$th hatchet is the the very least factorial many of $n$.

No ... Wait ... It"s $45$. The $n$th term is the biggest fourth-power-free divisor the $n!$.

Hold on ... :)

Probably the answer they"re feather for, though, is $6! = 720$. But there are lots of various other justifiable answers! After some testing I discovered that these numbers space being multiply by their corresponding number in the sequence.

For example:

1 x 2 = 22 x 3 = 66 x 4 = 2424 x 5 = 120Which would typical the next number in the sequence would be

120 x 6 = 720and for this reason on and so forth.

Edit: thanks to
GEdgar in the comments for helping me do pretty cool discovery around these numbers. The totals are additionally made increase of multiplying each number as much as that existing count.

For Example:

2! = 2 x 1 = 23! = 3 x 2 x 1 = 64! = 4 x 3 x 2 x 1 = 245! = 5 x 4 x 3 x 2 x 1 = 1206! = 6 x 5 x 4 x 3 x 2 x 1 = 720 The next number is 720.

The succession is the factorials:

1 2 6 24 120 = 1! 2! 3! 4! 5!

6! = 720.

(Another way to think of it is each term is the term prior to times the following counting number.

See more: Red Dead Redemption 2 Look On My Works, American Dreams

T0 = 1; T1 = T0 * 2 = 2; T2 = T1 * 3 = 6; T3 = T2 * 4 = 24; T4 = T3 * 5 = 120; T5 = T4 * 6 = 720. $\begingroup$ it's yet done. You re welcome find one more answer , a small bit initial :) perhaps with the amount of the number ? note also that it begins with 1 2 and ends v 120. Perhaps its an opportunity to concatenate and add zeroes. An excellent luck $\endgroup$

## Not the prize you're feather for? Browse other questions tagged sequences-and-series or ask your very own question.

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