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You are watching: 1 to 2 to 6 to 24 to 120
I was playing v No Man"s Sky when I ran right into a collection of numbers and also was asked what the following number would be.
$$1, 2, 6, 24, 120$$
This is for a terminal assess code in the game no mans sky. The 3 choices they give are; 720, 620, 180
The following number is $840$. The $n$th term in the succession is the the smallest number with $2^n$ divisors.
Er ... The next number is $6$. The $n$th ax is the the very least factorial many of $n$.
No ... Wait ... It"s $45$. The $n$th term is the best fourth-power-free divisor that $n!$.
Hold on ... :)
Probably the prize they"re spring for, though, is $6! = 720$. However there room lots of other justifiable answers!
After some trial and error I found that these numbers space being multiplied by their equivalent number in the sequence.
1 x 2 = 22 x 3 = 66 x 4 = 2424 x 5 = 120Which would typical the next number in the sequence would certainly be
120 x 6 = 720and therefore on and also so forth.
Edit: many thanks to
GEdgar in the comments for helping me make pretty cool discovery about these numbers. The totals are also made up of multiplying each number as much as that existing count.
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 means to think of that is every term is the term before times the following counting number.
See more: Byzantine Scholars Produced Many Notable Books Especially In The Field Of A
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 however done. Please find another answer , a tiny bit original :) perhaps with the amount of the digits ? note additionally that it begins with 1 2 and ends through 120. Maybe its an possibility to concatenate and add zeroes. Good luck $\endgroup$
Not the prize you're spring for? Browse other questions tagged sequences-and-series or ask your very own question.
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