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

%I #6 Feb 17 2018 20:01:55

%S 1,1,2,1,3,3,6,5,8,7,9,9,12,11,14,13,15,15,18,17,20,19,21,21,24,23,26,

%T 25,27,27,30,29,32,31,33,33,36,35,38,37,39,39,42,41,44,43,45,45,48,47,

%U 50,49,51,51,54,53,56,55,57,57,60,59,62,61,63,63,66,65,68,67

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

%H Clark Kimberling, <a href="/A214920/b214920.txt">Table of n, a(n) for n = 2..1000</a>

%e Write x#y if x|y is false; then 21#123, 13#199, 8#322, 5#521, 3|843 so a(9) = 5.

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

%Y Cf. A214919.

%K nonn,easy

%O 2,3

%A _Clark Kimberling_, Jul 30 2012