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


I to be playing with No Man"s Sky as soon as I ran right into a series of numbers and was asked what the next number would be.

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

This is because that a terminal assess password in the video game no man sky. The 3 options they give are; 720, 620, 180


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The following number is $840$. The $n$th ax in the succession is the the smallest number with $2^n$ divisors.

Er ... The following number is $6$. The $n$th term is the least factorial multiple of $n$.

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

Hold ~ above ... :)

Probably the answer they"re feather for, though, is $6! = 720$. But there space lots of various other justifiable answers!


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After some experimentation I discovered that these numbers room being multiply by their equivalent number in the sequence.

For example:

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

120 x 6 = 720and so on and also so forth.

Edit: many thanks to
GEdgar in the comments because that helping me make pretty cool discovery around these numbers. The totals are also made up of multiplying each number approximately 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

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The following 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 that is each term is the term prior to times the next counting number.

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T0 = 1; T1 = T0 * 2 = 2; T2 = T1 * 3 = 6; T3 = T2 * 4 = 24; T4 = T3 * 5 = 120; T5 = T4 * 6 = 720.


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$\begingroup$ it's yet done. Please find an additional answer , a small bit initial :) maybe with the sum of the digits ? note also that it begins with 1 2 and also ends v 120. Possibly its an possibility to concatenate and include zeroes. Great luck $\endgroup$

Not the prize you're looking for? Browse various other questions tagged sequences-and-series or asking your own question.


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