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A267806 a(0) = a(1) = 1; for n>1, a(n) = (a(n-1) mod 2) + a(n-2).

%I A267806 #17 May 02 2023 11:16:43
%S A267806 1,1,2,1,3,2,3,3,4,3,5,4,5,5,6,5,7,6,7,7,8,7,9,8,9,9,10,9,11,10,11,11,
%T A267806 12,11,13,12,13,13,14,13,15,14,15,15,16,15,17,16,17,17,18,17,19,18,19,
%U A267806 19,20,19,21,20,21,21,22,21,23,22,23,23,24,23,25,24,25,25
%N A267806 a(0) = a(1) = 1; for n>1, a(n) = (a(n-1) mod 2) + a(n-2).
%H A267806 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,1,0,-1).
%F A267806 From _Bruno Berselli_, Jan 21 2016: (Start)
%F A267806 G.f.: (1 + x + x^2 - x^3)/((1 + x)*(1 - x)^2*(1 + x + x^2)).
%F A267806 a(n) = a(n-2) + a(n-3) - a(n-5) for n>4.
%F A267806 a(n) = floor((n + 2)/3) + (1 + (-1)^n)/2. (End)
%F A267806 a(n) = A051274(n+2). - _R. J. Mathar_, May 02 2023
%t A267806 RecurrenceTable[{a[0] == a[1] == 1, a[n] == Mod[a[n - 1], 2] + a[n - 2]}, a, {n, 80}]
%t A267806 Table[Floor[(n + 2)/3] + (1 + (-1)^n)/2, {n, 0, 80}] (* or *) LinearRecurrence[{0, 1, 1, 0, -1}, {1, 1, 2, 1, 3}, 80] (* _Bruno Berselli_, Jan 21 2016 *)
%o A267806 (PARI) a=vector(100); for(n=1, #a, if(n<3, a[n]=1, a[n]=a[n-1]%2+a[n-2])); a \\ _Colin Barker_, Jan 22 2016
%Y A267806 Cf. A000045, A267807, A267808, A267809.
%K A267806 nonn,easy
%O A267806 0,3
%A A267806 _José María Grau Ribas_, Jan 20 2016
%E A267806 Edited by _Bruno Berselli_, Jan 21 2016.