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a(0) = 5, a(1) = 8; for n >1, a(n)=(a(n-1)+a(n-2))/3^n, where 3^n is the highest power of 3 dividing a(n-1)+a(n-2).
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%I #12 Mar 13 2024 08:30:57

%S 5,8,13,7,20,1,7,8,5,13,2,5,7,4,11,5,16,7,23,10,11,7,2,1,1,2,1,1,2,1,

%T 1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,

%U 2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2,1,1,2

%N a(0) = 5, a(1) = 8; for n >1, a(n)=(a(n-1)+a(n-2))/3^n, where 3^n is the highest power of 3 dividing a(n-1)+a(n-2).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,1).

%F a(3n-1)=2, a(3n)=1, a(3n+1)=1 for n>=8. - _Sascha Kurz_, Jan 04 2003

%t Join[{5, 8, 13, 7, 20, 1, 7, 8, 5, 13, 2, 5, 7, 4, 11, 5, 16, 7, 23, 10, 11, 7},LinearRecurrence[{0, 0, 1},{2, 1, 1},79]] (* _Ray Chandler_, Aug 25 2015 *)

%Y Cf. A038502, A078414.

%K nonn,easy

%O 0,1

%A _Yasutoshi Kohmoto_, Dec 28 2002

%E More terms from _Sascha Kurz_, Jan 04 2003