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**Mcbattle****Member**- Registered: 2014-01-29
- Posts: 5

So after I failed math last semester I got kicked into general math. But I already knew the concepts (I got a 94 after paying attention for less than 15 minutes over the course of 4 weeks) so I did what any other board 14 year does... screw around. I knew that if you take the remainder after dividing the Fibonacci numbers you see a pattern. However I noticed a pattern in the length of the patterns, but only when it was divided by certain numbers. The numbers in question would be 2^n for example:

2^1=2 The sequence is 3 numbers: 0,1,1

2^2=4 The sequence is 6 numbers: 0,1,1,2,3,1

2^3=8 The sequence is 12 numbers: 0,1,1,2,3,5,0,5,5,2,7,1

The pattern I noticed is that the length of the sequence is equal to

I have confirmed the same thing with 16 and 32 while in math by writing a program on my calculator. I was unable to go higher due to technical reasons though. I plan on looking into this later and seeing if it does continue.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,417

hi Mcbattle

Welcome to the forum.

Sorry, I'm not following you. Would you fill in more details for an old brain.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Mcbattle****Member**- Registered: 2014-01-29
- Posts: 5

Yes that is it. Or to make it easier it is the sum of the last two numbers in the sequence.

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I posted your query on another forum: http://www.artofproblemsolving.com/Foru … 8&t=573625.

As of now the thread has had 46 views but no reply.

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Ive been given this link (PDF):

**233** books currently added on Goodreads

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**iamaditya****Member**- From: Planet Mars
- Registered: 2016-11-15
- Posts: 790

In Fibbonacci sequence the next term is found by adding the previous 2 terms. But if we add all the previous terms before it to get the next no. we will get

1,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192..........

which is nothing but powers of 2.So,it is actually

2^0,2^0,2^1,2^2,2^3,2^4,2^5,2^6.........

*Last edited by iamaditya (2016-11-30 23:05:06)*

Practice makes a man perfect.

There is no substitute to hard work

All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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**zahlenspieler****Member**- Registered: 2018-04-12
- Posts: 7

Mcbattle wrote:

Yes that is it. Or to make it easier it is the sum of the last two numbers in the sequence.

Hi Mcbattle, I've just finished an inductive proof of your claim: Let

, and if . Then, and

for all integers .

(the 2nd statement is needed to complete the induction step.)

Furthermore, you need

I guess the easiest way is to use the Euler-Binnet formula; with a little more work, it can be proved without it.

Regards,

zahlenspieler

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