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A267808 a(0) = a(1) = 1; for n>1, a(n) = (a(n-1) mod 4) + a(n-2).

%I A267808 #25 Jan 22 2016 11:26:08
%S A267808 1,1,2,3,5,4,5,5,6,7,9,8,9,9,10,11,13,12,13,13,14,15,17,16,17,17,18,
%T A267808 19,21,20,21,21,22,23,25,24,25,25,26,27,29,28,29,29,30,31,33,32,33,33,
%U A267808 34,35,37,36,37,37,38,39,41,40,41,41,42,43,45,44,45,45,46,47,49
%N A267808 a(0) = a(1) = 1; for n>1, a(n) = (a(n-1) mod 4) + a(n-2).
%H A267808 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,1,-1).
%F A267808 From _Bruno Berselli_, Jan 21 2016: (Start)
%F A267808 G.f.: (1 + x^2 + x^3 + 2*x^4 - x^5) / ((1 + x)*(1 - x)^2*(1 - x + x^2)*(1 + x + x^2)).
%F A267808 a(n) = a(n-1) + a(n-6) - a(n-7) for n>6.
%F A267808 a(n) = n - floor((n+1)/3) + ((-1)^n - (-1)^(floor(n/3)))/2 + 1. (End)
%t A267808 RecurrenceTable[{a[0] == a[1] == 1, a[n] == Mod[a[n - 1], 4] + a[n - 2]}, a, {n, 70}]
%t A267808 Table[n - Floor[(n + 1)/3] + ((-1)^n - (-1)^Floor[n/3])/2 + 1, {n, 0, 70}] (* or *) LinearRecurrence[{1, 0, 0, 0, 0, 1, -1}, {1, 1, 2, 3, 5, 4, 5}, 80] (* _Bruno Berselli_, Jan 21 2016 *)
%o A267808 (PARI) a=vector(100); for(n=1, #a, if(n<3, a[n]=1, a[n]=a[n-1]%4+a[n-2])); a \\ _Colin Barker_, Jan 22 2016
%Y A267808 Cf. A000045, A267806, A267807, A267809.
%K A267808 nonn,easy
%O A267808 0,3
%A A267808 _José María Grau Ribas_, Jan 20 2016
%E A267808 Edited by _Bruno Berselli_, Jan 21 2016.