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A274845 a(0)=1, a(1)=0, a(4n+2) = a(4n+3) = a(4n+5) = (4^(n+1) +(-1)^n)/5, a(4n+4) = (2*4^(n+1) -3*(-1)^n)/5.

%I A274845 #13 Aug 08 2016 23:50:01
%S A274845 1,0,1,1,1,1,3,3,7,3,13,13,25,13,51,51,103,51,205,205,409,205,819,819,
%T A274845 1639,819,3277,3277,6553,3277,13107,13107,26215,13107,52429,52429,
%U A274845 104857,52429,209715,209715,419431,209715,838861,838861,1677721,838861,3355443
%N A274845 a(0)=1, a(1)=0, a(4n+2) = a(4n+3) = a(4n+5) = (4^(n+1) +(-1)^n)/5, a(4n+4) = (2*4^(n+1) -3*(-1)^n)/5.
%C A274845 Antidiagonals of the array in A274613 written as a triangle:
%C A274845 1,
%C A274845 0, 1/2,
%C A274845 0, 1/2, 1/4,
%C A274845 0,   0, 1/2, 1/8,
%C A274845 0,   0, 1/4, 3/8, 1/16,
%C A274845 ... .
%C A274845 a(n) is the numerators of the antidiagonal sums i.e. 1, 0, 1/2, 1/2, 1/4, 1/2, 3/8, 3/8, 7/16, 3/8, 13/32, 13/32, 25/64, 13/32, ... = a(n)/b(n).
%C A274845 The denominators b(n) are A173300(n).
%C A274845 a(0)+a(1) = 1, a(4n+2) +a(4n+3) +a(4n+4) +a(4n+5) = 4, 16, 64, 256, ... .
%H A274845 Colin Barker, <a href="/A274845/b274845.txt">Table of n, a(n) for n = 0..1000</a>
%H A274845 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 3, 0, 0, 0, 4).
%F A274845 a(4n) = A102900(n), a(4n+1) = A015521(n), a(4n+2) = a(4n+3) = A015521(n+1).
%F A274845 a(n) = 3*a(n-4) + 4*a(n-8). - _G. C. Greubel_, Jul 08 2016
%F A274845 G.f.: (1+x^2+x^3-2*x^4+x^5) / ((1-2*x^2)*(1+2*x^2)*(1+x^4)). - _Colin Barker_, Jul 22 2016
%t A274845 LinearRecurrence[{0,0,0,3,0,0,0,4}, {1,0,1,1,1,1,3,3}, 50] (* _G. C. Greubel_, Jul 08 2016 *)
%o A274845 (PARI) Vec((1+x^2+x^3-2*x^4+x^5)/((1-2*x^2)*(1+2*x^2)*(1+x^4)) + O(x^60)) \\ _Colin Barker_, Jul 22 2016
%Y A274845 Cf. A000302, A102900, A015521, A173300, A274613.
%K A274845 nonn,easy
%O A274845 0,7
%A A274845 _Paul Curtz_, Jul 08 2016
%E A274845 More terms from _Colin Barker_, Jul 22 2016