Fibonacci series

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 

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Fibonacci series  ggauny@live.fr  11062015, 11:59 AM
RE: Fibonacci series  Paul Dale  11062015 12:17 PM
RE: Fibonacci series  ggauny@live.fr  11062015, 12:29 PM

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