#### Posers and Puzzles

Germany

- Joined
- 27 Oct '08
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- 3118

What is the next number in the following sequence?

0

1

2

2.60121894356579510020490322708104361119152187501694... x 10^1746 (approximation)

at home

- Joined
- 09 Jun '07
- Moves
- 46110

*Originally posted by KazetNagorra*

**What is the next number in the following sequence?
**

0

1

2

2.60121894356579510020490322708104361119152187501694... x 10^1746 (approximation)

42 (approximation)

Germany

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- 27 Oct '08
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*Originally posted by wolfgang59*

**42 (approximation)**

42 is

*not* the answer!

- Joined
- 26 Apr '03
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- 26588

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.

podunk, PA

- Joined
- 10 Dec '06
- Moves
- 7733

*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

Germany

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4!!!! is the correct answer.

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

- Joined
- 11 Nov '05
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- 43938

*Originally posted by KazetNagorra*

**4!!!! is the correct answer.
**

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!

podunk, PA

- Joined
- 10 Dec '06
- Moves
- 7733

*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.

ðŸ™‚

Germany

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- 27 Oct '08
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- 3118

*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.

- Joined
- 26 Apr '03
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- 26588

*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.