# Number sequence

KazetNagorra
Posers and Puzzles 03 Mar '10 13:11
1. 03 Mar '10 13:113 edits
What is the next number in the following sequence?

0
1
2
2.60121894356579510020490322708104361119152187501694... x 10^1746 (approximation)
2. wolfgang59
Mr. Wolf
03 Mar '10 22:22
Originally posted by KazetNagorra
What is the next number in the following sequence?

0
1
2
2.60121894356579510020490322708104361119152187501694... x 10^1746 (approximation)
42 (approximation)
3. 04 Mar '10 07:17
Originally posted by wolfgang59
42 (approximation)
4. 04 Mar '10 20:00
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.
5. joe shmo
Strange Egg
04 Mar '10 20:08
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
6. 04 Mar '10 20:56

Or written alternatively: 10^(10^(10^25.16114896940657)).
7. 04 Mar '10 21:10
Originally posted by KazetNagorra

Or written alternatively: 10^(10^(10^25.16114896940657)).
A good one! ðŸ™‚

I just thought the problem as silly at first, but now I enjoy the problem, and its solution!
8. joe shmo
Strange Egg
05 Mar '10 00:06
Originally posted by KazetNagorra

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.

ðŸ™‚
9. 05 Mar '10 07:23
Originally posted by joe shmo
Is there a general series definition for this? I tried to develop one but my head exploded.

ðŸ™‚
I have no idea, this is just what Wolfram Alpha gave me.
10. 09 Mar '10 01:03
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.