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a(n) = a(a(n-3)) + a(n-a(n-2)) with a(1) = a(2) = a(3) = a(4) = 1, a(5) = 2, a(6) = 5.
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%I #22 Apr 20 2022 17:22:54

%S 1,1,1,1,2,5,3,2,7,3,2,10,3,2,13,3,2,16,3,2,19,3,2,22,3,2,25,3,2,28,3,

%T 2,31,3,2,34,3,2,37,3,2,40,3,2,43,3,2,46,3,2,49,3,2,52,3,2,55,3,2,58,

%U 3,2,61,3,2,64,3,2,67,3,2,70,3,2,73,3,2,76,3,2,79,3,2,82,3,2,85,3,2,88

%N a(n) = a(a(n-3)) + a(n-a(n-2)) with a(1) = a(2) = a(3) = a(4) = 1, a(5) = 2, a(6) = 5.

%C A quasi-periodic solution to the recurrence a(n) = a(a(n-3)) + a(n-a(n-2)).

%H Colin Barker, <a href="/A302130/b302130.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).

%F From _Colin Barker_, Jun 20 2018: (Start)

%F G.f.: x*(1 + x + x^2 - x^3 + 3*x^5 + 2*x^6 - x^7 - 2*x^8 - 2*x^9 + x^11) / ((1 - x)^2*(1 + x + x^2)^2).

%F a(n) = 2*a(n-3) - a(n-6) for n>10.

%F (End)

%e a(3*k-2) = 3, a(3*k-1) = 2, a(3*k) = 3*k - 2 for k > 2.

%t LinearRecurrence[{0,0,2,0,0,-1},{1,1,1,1,2,5,3,2,7,3,2,10},100] (* _Harvey P. Dale_, Apr 20 2022 *)

%o (PARI) a=vector(99); a[1]=a[2]=a[3]=a[4]=1;a[5]=2;a[6]=5;for(n=7, #a, a[n] = a[a[n-3]]+a[n-a[n-2]]); a

%o (PARI) Vec(x*(1 + x + x^2 - x^3 + 3*x^5 + 2*x^6 - x^7 - 2*x^8 - 2*x^9 + x^11) / ((1 - x)^2*(1 + x + x^2)^2) + O(x^80)) \\ _Colin Barker_, Jun 20 2018

%o (GAP) a:=[1,1,1,1,2,5];; for n in [7..100] do a[n]:=a[a[n-3]]+a[n-a[n-2]]; od; a; # _Muniru A Asiru_, Jun 26 2018

%Y Cf. A244477.

%K nonn,easy

%O 1,5

%A _Altug Alkan_, Jun 20 2018