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a(n) is the least m > 0 such that Fibonacci(n-m) divides Fibonacci(n+2+m).
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%I #6 Feb 17 2018 20:01:44

%S 1,1,2,1,4,3,2,4,8,3,10,6,4,7,14,5,16,9,6,10,20,7,14,12,8,13,26,9,28,

%T 15,10,16,20,11,34,18,12,19,38,13,40,21,14,22,44,15,34,24,16,25,50,17,

%U 32,27,18,28,56,19,58,30,20,31,38,21,64,33,22,34,68,23,70,36

%N a(n) is the least m > 0 such that Fibonacci(n-m) divides Fibonacci(n+2+m).

%H Clark Kimberling, <a href="/A214918/b214918.txt">Table of n, a(n) for n = 2..1000</a>

%e Write x#y if x|y is false; then 55#377, 34#610, 21|987 so a(11) = 3.

%t Table[m = 1; While[! Divisible[Fibonacci[n+2+m],Fibonacci[n-m]], m++]; m, {n, 2, 100}]

%Y Cf. A214917.

%K nonn,easy

%O 2,3

%A _Clark Kimberling_, Jul 30 2012