9^(9^9)

[Batalov]

The real poll should have been about semantics:

What is the value of 9^9^9?
* 9^81
* 9^(9^9)
* 9 if ^ is XOR operation :-)

Hans Landa: Is that the way you say it? "That's a bingo?"
Aldo Raine: You just say "bingo".
Hans Landa: Ah! BINGO! How fun! But, I digress. Where were we?

[lavalamp]

Originally Posted by Raman

9^9[SUP]9[/SUP] = 9^387420489 is being the largest number that can be written with 3 digits.

^[SUP]99[/SUP]9

[R.D. Silverman]

Originally Posted by Batalov

9!^9!^9! seems to be slightly larger.

If you are going to allow functions, there is no limit to the size
of the number. eg A(9^9, 9) where A is Ackerman's Function is
far larger, as is BB(9^9^9) where BB is the Busy Beaver Function.
Or we could use Goodstein sequences, or other functions in the
fast growing hierarchy. ......

'nuff said???

[R.D. Silverman]

Originally Posted by m_f_h

Well, in a case like this one, it is not *so* bad to write 9^(9^9) to make things 100% unambiguous. .

They are unambiguous. Order of operations in an exponential tower
is well-defined. There is no need for parens.

[Raman]

Originally Posted by lavalamp

^[SUP]99[/SUP]9

This is being the tetration (aka hyperexponentiation) operation
The last few digits rather converges for repeatly iteratively applying this function composition as such

Reminding me rather for this Ackermann function, as such
The function value Ackermann(9,9⁹) is rather going to be far larger enough, as such, man

whereby

A(0,n) = n+1
A(1,n) = 2+(n+3)-3
A(2,n) = 2*(n+3)-3
A(3,n) = 2ⁿ⁺³-3 = 2↑(n+3)-3
A(4,n) = ⁿ⁺³2 - 3 = 2↑↑(n+3)-3
...
A(m,n) = 2↑^n-2(n+3)-3

[retina]

Originally Posted by Raman

A(m,n) = 2↑^n-2(n+3)-3

Where is m on the RHS?

[Raman]

Originally Posted by retina

Where is m on the RHS?

A(m,n) = 2↑^m-2(n+3)-3


Thanks for pointing it out,
that error to me out

I was during that time period
rushing off to post message number # 300000
some mistake had happened meanwhile
rather within
away that way

For example, A(4,2) = 2⁶⁵⁵³⁶-3
(it is being the 19729 digit number, as such, as since)

[science_man_88]

Originally Posted by Raman

This is being the tetration (aka hyperexponentiation) operation

I with a little reading I'd read it as a pentation:

^[SUP]99[/SUP]9 = ⁹(9^{9[SUP]9[SUP]9[SUP]9[SUP]9[SUP]9[SUP]9[SUP]9[SUP]9}[/SUP][/SUP][/SUP][/SUP][/SUP][/SUP][/SUP][/SUP]) = 9³ 9

[Stargate38]

The last 3 digits of 9^(9^9) are 289. I used Pari-gp to calculate 9^489 mod 1000 which is the same as 9^1489 mod 1000, 9^2489 mod 1000, etc. because 9^1000 mod 1000 = 1 and 9^489 mod 1000 is 289. I also voted in the poll.

[retina]

Originally Posted by Stargate38

The last 3 digits of 9^(9^9) are 289. I used Pari-gp to calculate 9^489 mod 1000 which is the same as 9^1489 mod 1000, 9^2489 mod 1000, etc. because 9^1000 mod 1000 = 1 and 9^489 mod 1000 is 289. I also voted in the poll.

I thought it was simpler than that. 9⁵⁰ mod 1000 ≡ 1, and 9⁹ mod 50 ≡ 39, leads to 9³⁹ mod 1000 ≡ 289.

[LaurV]

So much talking for nothing...

(11:39:36) gp > lift((x=Mod(9,1000))^lift(Mod(9,znorder(x))^9))
%8 = 289
(11:39:36) gp > ##
  ***   last result computed in 0 ms.
(11:39:37) gp >

edit: more digits:

(11:43:54) gp > digits=100; lift((x=Mod(9,10^digits))^lift(Mod(9,znorder(x))^9))
%11 = 2617004315060225040660196165699439754361026855266374036682906190174923494324178799359681422627177289
(11:43:55) gp > ##
  ***   last result computed in 12 ms.
(11:44:06) gp >