Fibonacci series
11-06-2015, 12:17 PM
Post: #2
 Paul Dale Senior Member Posts: 1,839 Joined: Dec 2013
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
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 Messages In This Thread Fibonacci series - ggauny@live.fr - 11-06-2015, 11:59 AM RE: Fibonacci series - Paul Dale - 11-06-2015 12:17 PM RE: Fibonacci series - ggauny@live.fr - 11-06-2015, 12:29 PM

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