Number sequence

What is the next number in the following sequence?

0
1
2
2.60121894356579510020490322708104361119152187501694... x 10^1746 (approximation)

Originally posted by KazetNagorra
What is the next number in the following sequence?

0
1
2
2.60121894356579510020490322708104361119152187501694... x 10^1746 (approximation)

42 (approximation)

Originally posted by wolfgang59
42 (approximation)

42 is not the answer!

Hmm, the answer seems to depend on whether the sequence is:

1!
2!!
3!!!

or
1!!!
2!!!
3!!!

which give the same answers up to the third value, which is all we have.

Originally posted by iamatiger
Hmm, the answer seems to depend on whether the sequence is:

1!
2!!
3!!!

or
1!!!
2!!!
3!!!

which give the same answers up to the third value, which is all we have.

the 0 in the sequence gives the solution

cant be 0!!! =1, thus the first term must be zero

so

0
1!
2!!
3!!!
4!!!!

nice catch

4!!!! is the correct answer.

Or written alternatively: 10^(10^(10^25.16114896940657)).

Originally posted by KazetNagorra
4!!!! is the correct answer.

Or written alternatively: 10^(10^(10^25.16114896940657)).

A good one! 🙂

I just thought the problem as silly at first, but now I enjoy the problem, and its solution!

Originally posted by KazetNagorra
4!!!! is the correct answer.

Or written alternatively: 10^(10^(10^25.16114896940657)).

Is there a general series definition for this? I tried to develop one but my head exploded.

🙂

Originally posted by joe shmo
Is there a general series definition for this? I tried to develop one but my head exploded.

🙂

I have no idea, this is just what Wolfram Alpha gave me.

Originally posted by KazetNagorra
I have no idea, this is just what Wolfram Alpha gave me.

Looking at: http://en.wikipedia.org/wiki/Factorial

Wolfram may be using the equation given under "multifactorials", which uses k as "the number of exclamation marks.