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Very Big Number
By Vladimir Orlovsky on Friday, February 19, 1999 - 11:44 am:

Hello.
I like to know, where I can find Info about.....

f(4) = 432 = 49 = 262144
f(5) = 5432 = 5262144 ~~ 9×10183222

f(6) = ? VERY BIG NUMBER ?

Thanks.

By David Sanders (Dps24) on Thursday, March 25, 1999 - 10:42 pm:

Hi Vladimir,

I'm afraid I've never seen anyone discuss this function, so I can't tell you where to look. One of the problems, as you say, is that it does get VERY big VERY quickly.

But we can get an idea of the size of the number in terms of number of digits by using logarithms (I hope you're familiar with these -- I presume it's how you found your value for f(5)?)

As you say, f(4) = 262144, and your calculator can tell you that. But if you try f(5) on a calculator, it will probably complain!!

But we know that:

log(f(5)) = log(5262144) = 262144 × log(5)
= 262144 × 0.69897
= 183230.793

(I am using 'common logarithms', i.e. log to the base 10, here.)

So f(5) is about 6.21×10183230
(so I disagree with your answer).

How did I get 6.21?

Well, f(5) = 10log(f(5)), by definition of log.
So, f(5) = 10183230.793
= 100.793 × 10183230
= 6.21 × 10183230

Note that this means that there are 183231 digits in the answer for f(5).

We can repeat this process to find f(6):

log(f(6)) = (log 6) × f(5)
= 0.778 × f(5)

But f(5) is rather large! This means that the NUMBER OF DIGITS in the answer is about 100000000000000000000000000000000000000000000000000.........000000, where the number of digits in THIS is 183231.

I don't know about you, but I think this number is just ridiculously big.

So in a way, I'm afraid f isn't a very interesting function, because there's no way you can get your head round how fast it grows!!


David.

By Anonymous on Thursday, October 28, 1999 - 05:33 pm:

Very Big Numbers:
================
Vladimir defined a new number type in the following way:
f(1) = 1
f(n) = nf(n-1) for any n > 1 where n is integer

Some Computation:
================
f(1) = 1
f(2) = 21 = 2
f(3) = 32 = 9
f(4) = 49 = 262144
f(5) = 5262144 = It is very large number containing 183231 digits
so giving here the entire number is wastage of space.
Its approximate value is
6.206069878660874470748320557284679309194219265199117173177383×10 183230

As the number grows very very fast, it is not possible to go beyond
this.

However, we can compute f(n) modulo x where x is some integer. Before
going ahead, let us understand what is modulo. It means the remainder
if f(n) is divided by x. For example, 17 modulo 7 is 3, when 17 is
divided by 7, the remainder will be 3.

Below, we have computed f(2) to f(50) for Modulo 123456789. This is
done in gp-pari. The entire program is of one line. The computation
is almost instantaneous.

// Program
? b=123456789;s=Mod(1,b);for(i=2,50,s=Mod(i,b)^lift(s);print(i," ",s))
// Result
2 Mod(2, 123456789)
3 Mod(9, 123456789)
4 Mod(262144, 123456789)
5 Mod(16317634, 123456789)
6 Mod(49450158, 123456789)
7 Mod(81457021, 123456789)
8 Mod(46456721, 123456789)
9 Mod(23229306, 123456789)
10 Mod(71403760, 123456789)
11 Mod(7740250, 123456789)
12 Mod(99701091, 123456789)
13 Mod(56086183, 123456789)
14 Mod(84465257, 123456789)
15 Mod(118437120, 123456789)
16 Mod(7362865, 123456789)
17 Mod(78259193, 123456789)
18 Mod(34137045, 123456789)
19 Mod(69135787, 123456789)
20 Mod(101388530, 123456789)
21 Mod(21627414, 123456789)
22 Mod(111339586, 123456789)
23 Mod(21159490, 123456789)
24 Mod(113848326, 123456789)
25 Mod(122242069, 123456789)
26 Mod(53719928, 123456789)
27 Mod(106512102, 123456789)
28 Mod(83638252, 123456789)
29 Mod(117497815, 123456789)
30 Mod(46783287, 123456789)
31 Mod(36512497, 123456789)
32 Mod(70363760, 123456789)
33 Mod(46473750, 123456789)
34 Mod(113844079, 123456789)
35 Mod(115577810, 123456789)
36 Mod(35529876, 123456789)
37 Mod(67679353, 123456789)
38 Mod(35140088, 123456789)
39 Mod(80813313, 123456789)
40 Mod(93133945, 123456789)
41 Mod(18161087, 123456789)
42 Mod(106605414, 123456789)
43 Mod(109481149, 123456789)
44 Mod(6486578, 123456789)
45 Mod(41640516, 123456789)
46 Mod(30969271, 123456789)
47 Mod(87982679, 123456789)
48 Mod(49260519, 123456789)
49 Mod(87691825, 123456789)
50 Mod(65411492, 123456789)
Good bye!

Final Note:
==========
f(50) Modulo 123456789 is 65411492.
This means if we compute a number:
504948....21
and then divide it by 123456789 then remainder will be 65411492.

It is practically impossible to compute even f(6) using the best
computer in the world but we are allowed to do Modulo arithmetic
on it [even on f(50) or f(1000)] using any ordinary computer. Is
it not fantastic?

P L Patodia, Bangalore, India.

By Neil Morrison (P1462) on Saturday, January 8, 2000 - 01:31 pm:

Has anybody heard of Graham's number? It has a similarity to the numbers you are talking about. It has a special notation. I may write more about it later

By The Editor:

There is some information about Graham's number on this site or this site. Please don't believe anything about it being the world's largest number, though!