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Inequalities for large integers


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14 replies to this topic

#1 Rainman

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Posted 29 January 2014 - 12:49 PM

These problems use Knuth's up-arrow notation (http://en.wikipedia....-arrow_notation), with ^ being the up-arrow.

 

1. Find the smallest value for n, such that 1000^^n > 12^^(n+1).

 

2. We define the sequence g, by assigning g(1) = 3^^^^3, and g(n+1) = 3^^^...^3, where the number of up-arrows is g(n). A famously large number is Graham's number = g(64). Which is larger, Graham's number or 2^^^...^2, where the number of up-arrows is Graham's number?


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#2 bonanova

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Posted 02 February 2014 - 11:17 AM

Aaaargh. Up arrows are famously non intuitive to think about. I'm thinking nonetheless.


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#3 Rainman

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Posted 07 February 2014 - 09:36 PM

Spoiler for general hint

Spoiler for hint for puzzle 1

Spoiler for hint for puzzle 2


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#4 phil1882

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Posted 08 February 2014 - 09:06 PM

Spoiler for

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#5 phil1882

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Posted 08 February 2014 - 09:26 PM

Spoiler for minor follow up

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#6 phil1882

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Posted 08 February 2014 - 09:51 PM

whoops....

Spoiler for big correction

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#7 Rainman

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Posted 08 February 2014 - 11:52 PM

Actually, just one up-arrow is the same as exponentiation.

3^3 = 33 = 27

3^^3 = 3^(3^3) = 3^27 = 7625597484987

3^^^3 = 3^^(3^^3) = 3^^(7625597484987) = 3^3^3^3^3...^3, where the number of threes is 7,625,597,484,987.

 

So 12^^2 = 12^12 = 8916100448256, which is much larger than 1000^^1 = 1000. The inequality doesn't hold for n=1.

Generally, 1000^^n is 1000^1000^1000^...^1000, where the number of 1000s is n.

 

You solved puzzle 2 :) 2^^^....^2 always equals 4, no matter how many up-arrows you have.


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#8 Rainman

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Posted 23 February 2014 - 07:50 PM

Spoiler for image for puzzle 1

Spoiler for further hint


Edited by Rainman, 23 February 2014 - 07:55 PM.

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#9 phil1882

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Posted 23 February 2014 - 10:58 PM

hmmm.

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#10 Rainman

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Posted 24 February 2014 - 12:27 AM

12^12^12 should be interpreted as 12^(12^12) rather than (12^12)^12. So log(12^12^12) = log(12^(12^12)) = (12^12)*log(12) ~ 9 600 000 000 000.


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