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a(n) = (a(n-2) + a(n-3))/gcd(a(n-2), a(n-3)) with a(1) = a(2) = a(3) = 1.
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%I #9 Jun 19 2021 11:39:04

%S 1,1,1,2,2,3,2,5,5,7,2,12,9,7,7,16,2,23,9,25,32,34,57,33,91,30,124,

%T 121,77,245,18,46,263,32,309,295,341,604,636,945,310,527,251,27,778,

%U 278,805,528,1083,1333,537,2416,1870,2953,2143,4823,5096,6966,109

%N a(n) = (a(n-2) + a(n-3))/gcd(a(n-2), a(n-3)) with a(1) = a(2) = a(3) = 1.

%C A variation on A214551 with the second and third terms being added and divided by the greatest common divisor of the pair of numbers.

%H Reed Kelly, <a href="/A214646/b214646.txt">Table of n, a(n) for n = 1..1003</a>

%e a(6) = (a(4)+a(3))/gcd(a(4),a(3)) = (2+1)/1 = 3.

%e a(19) = (a(17)+a(16))/gcd(a(17),a(16)) = (2+16)/2 = 9.

%t DivGCDxy[n_, x_, y_, init_] := Module[{t, a, i}, t = init;

%t Do[AppendTo[t, (t[[-x]] + t[[-y]])/GCD[t[[-x]], t[[-y]]]], {n}];

%t t]; DivGCDxy[100, 2, 3, {1, 1, 1}]

%Y Cf. A214551, A000931.

%K nonn

%O 1,4

%A _Reed Kelly_, Jul 24 2012

%E NAME adapted to offset and b-file. - _R. J. Mathar_, Jun 19 2021