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A166578 a(n) = a(n-3) + 2^(n-4) with a(1) = 1, a(2) = 2, a(3) = 1.

%I A166578 #15 May 17 2016 21:06:20
%S A166578 1,2,1,2,4,5,10,20,37,74,148,293,586,1172,2341,4682,9364,18725,37450,
%T A166578 74900,149797,299594,599188,1198373,2396746,4793492,9586981,19173962,
%U A166578 38347924,76695845,153391690,306783380,613566757,1227133514,2454267028
%N A166578 a(n) = a(n-3) + 2^(n-4) with a(1) = 1, a(2) = 2, a(3) = 1.
%C A166578 a(n) and successive differences: 1,1,2,1,2,4,5,10,20,37; 0,1,-1,1,2,1,5,10,17,37; 1,-2,2,1,-1,4,5,7,20,37; -3,4,-1,-2,5,1,2,13,17,34; 7,-5,-1,7,-4,1,11,4,17,43; -12,4,8,-11,5,10,-7,13,26,25; Rows must be taken by pairs (companions because a(n)-a(n-3) alternatively gives A131577 and A011782 also companions). Note a(3n+2)=2*a(3n+1)=4*a(3n), n positive; see A113405. Sum of consecutive three terms of even rows gives 0,4,32,256.
%H A166578 G. C. Greubel, <a href="/A166578/b166578.txt">Table of n, a(n) for n = 0..1000</a>
%H A166578 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,1,-2).
%F A166578 For n > 4, a(n) = 2a(n-1) + a(n-3) - 2a(n-4).
%F A166578 a(3n) = (8^n - 8)/14 + 1, a(3n-1) = (8^n - 8)/28 + 2, a(3n-2) = (8^n - 8)/56 + 1.
%F A166578 G.f.: (1-3*x^2-x^3)/(1-2*x-x^3+2*x^4). - _Colin Barker_, Jan 25 2012
%t A166578 RecurrenceTable[{a[1]==1,a[2]==2,a[3]==1,a[n]==a[n-3]+2^(n-4)},a[n],{n,40}] (* or *) LinearRecurrence[{2,0,1,-2},{1,2,1,2},40] (* _Harvey P. Dale_, Sep 14 2011 *)
%o A166578 (PARI) Vec((1-3*x^2-x^3)/(1-2*x-x^3+2*x^4)+O(x^99)) \\ _Charles R Greathouse IV_, Jan 25 2012
%K A166578 nonn,easy
%O A166578 0,2
%A A166578 _Paul Curtz_, Oct 17 2009
%E A166578 Edited by _Charles R Greathouse IV_, Nov 04 2009