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Hi,

I have copiyied the good thread of Dieter on "tiny challenge of new year". It is a little different because of this.

We are going to 2016 year, and I was wonder if it is ways to find this 2016 in a Fibonnacci serie, or in a tribonano, Quadrari, quintina, sextina séries ?

As an example, see at followed :
For year 1999, the year of my little daughter is borned, I know this :

" 11 ; 52 ; 63 ; 115 ; 178 ; 293 ; 471 ; 764 ; 1235 ; 1999 "

A Fibonacci serie can biguin where we want.

It is doing the same but with 2016.

I am wished that you shall love this problem.

Cheerz

Me I have wrot for WP34s and Prime, with helped my son inginier in informatic

Gérard.
There is a trivial solution: 16; 2000; 2016 (okay, 0; 2016 is even more trivial). This leads easily to: 16; 1000; 1016; 2016. Can we do better?

More generally, the sequence starting from arbitrary a and b is: a; b; a+b; a+2b; 2a+3b; 3a+5b; 5a+8b; ...

Notice that the coefficients of a and b are consecutive Fibonacci numbers. Thus, any solution of the form FIB(n) * a + FIB(n+1) * b will work.

The Fibonacci numbers are: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, ...

This gives us more suitable sequences:

3; 670; 673; 1343; 2016 (using 2a+3b = 2016)
8; 247; 255; 502; 757; 1259; 2016 (using 5a + 8b = 2016)
0; 14; 14; 28; 42; 70; 112; 182; 294; 476; 770; 1246; 2016 (using 89a + 144b = 2016)


- Pauli
Hi,

I am inclined about you !

It is the solution of my son, I was thinking it is was very complicated, but for mathematicians....

I will reside in my range now,

Thanks for this so quickly respons, I am despite....

For me, without my son, it was not soluble !

Great thank,

Gérard. Sad
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