/sci/ - Science & Math

[Anonymous No.4813744]

can sci find out the billionth fibonacci number?

[sage No.4813749]

no

[Anonymous No.4813758]

>>4813755
upload it as a txt file

[Anonymous No.4813755]

Error: Field too long.

[Anonymous No.4813762]

anyone who knows a simple programming language can find out about it by themselves in about 30 seconds, without even using google

a simple for loop would suffice

[Anonymous No.4813760]

>>4813744
Use a recursive algorithm F(n)
F(1,000,000,000)

[Anonymous No.4813770]

>>4813762
I've been running it for 1 hour and it still hasn't stopped .

I would use the closed form expression but it gets inaccurate somewhere down the line.

[Anonymous No.4813771]

Simple.
f(1,000,000,000) = f(999,999,999) + f(999,999,998)

Give me another.

[Anonymous No.4813772]

>>4813771
find the 999,999,999th fibonacci number

[Anonymous No.4813774]

It has over 208 million digits. It is also divisible by 5. :)

[Anonymous No.4813776]

>>4813771
find the product of each number in the billionth fibonacci number

[Anonymous No.4813778]

>>4813749

Yes

<span class="math">F_{n} = \frac{(1+\sqrt{5})^{n} - (1-\sqrt{5})^{n}}{2^{n}\sqrt5}[/spoiler]

[Anonymous No.4813783]

>>4813772
f(999,999,999) = f(999,999,998) + f(999,999,997)

Give me another.

[Anonymous No.4813789]

>>4813778
>>4813778
>>4813778
>>4813778
>>4813778
>>4813778
>>4813778

[Anonymous No.4813799]

I can do it if you'll pay for my subscription to WolframAlpha Pro.

[Anonymous No.4813802]

>>4813799
start pro free trial

[None !IbCrowPiV2 No.4813855]

[Anonymous No.4813885]

>>4813855
>Windows
laughing_whores.jpg

[Anonymous No.4813887]

>>4813855

How long did that take?

[None !IbCrowPiV2 No.4813899]

>>4813885
Should I install gentoo?
>>4813887
5 minutes on overclocked i7 2600k.

[Anonymous No.4813925]

>>4813899

Apparently takes about five mins on anything,

Doesn't utilise all that cpu power

[None !IbCrowPiV2 No.4813950]

>>4813925
True, CPU was at 20% whole time. I don't think Mathematica utilizes hyper threading, which is stupid.

[Anonymous No.4813957]

>>4813887
It shouldn't take that long - after all, there's a closed form expression that involves the nth power, so it should be solvable in O(log(n))

[Anonymous No.4813999]

>>4813774

A text file containing the number is over 200mb

[Anonymous No.4814021]

You are all silly

Just look at >>4813778

[Anonymous No.4814028]

>>4813999
214,346,296 bytes

The exact length of the number

[Anonymous No.4814440]

Do you have python?
If you do use this function

import time
def fibonacci(n):
f1, f2 = 0, 1
x=1
for i in range(n):
print 'Step:'+str(x)+' - ' + str(f2)+'\n'
x+=1
time.sleep(0.2)
f1, f2 = f2, f1+f2

of course you should change or erase the time.sleep(x) line if you want to calculate large numbers, i just add it so you can see the numbers popping easier.

Depending on your computer it will take ages in big numbers though :)))

[Anonymous No.4814442]

>>4814440

damn identations got screwed up, just add a tab before the line after the :