Fibonacci series

11062015, 11:59 AM
Post: #1




Fibonacci series
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. Gérard. 

11062015, 12:17 PM
Post: #2




RE: Fibonacci series
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 

11062015, 12:29 PM
Post: #3




RE: Fibonacci series
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. Gérard. 

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