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%I #6 Jan 27 2017 13:22:05
%S 1,5,9,25,49,113,225,481,961,1985,3969,8065,16129,32513,65025,130561,
%T 261121,523265,1046529,2095105,4190209,8384513,16769025,33546241,
%U 67092481,134201345,268402689,536838145,1073676289,2147418113,4294836225,8589803521,17179607041
%N a(2n) = A060867(n+1), a(2n+1) = A092440(n+1).
%C In binary there is a pattern in how the zeros and ones appear:
%C a(0) = 01
%C a(1) = 101
%C a(2) = 1001
%C a(3) = 11001
%C a(4) = 110001
%C a(5) = 1110001
%C a(6) = 11100001
%C a(7) = 111100001
%C a(8) = 1111000001
%C a(9) = 11111000001
%C a(10) = 111110000001
%C a(11) = 1111110000001
%C a(12) = 11111100000001
%C a(13) = 111111100000001
%C a(14) = 1111111000000001
%C a(15) = 11111111000000001
%C Graphically, each term can be obtained by successively and alternately forming squares and centered squares as shown in the illustration.
%H Daniel Poveda Parrilla, <a href="/A276918/b276918.txt">Table of n, a(n) for n = 0..1000</a>
%H Daniel Poveda Parrilla, <a href="/A276918/a276918.gif">Illustration of initial terms</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-6,4).
%F a(n) = 1 + 2^(n+2) - 2^(1 + n/2) + (-1)^(n+1)*2^(1 + n/2) - 2^((n+1)/2) + (-1)^(n+2)*2^((n+1)/2).
%F a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4) for n>3.
%F G.f.: (-1-2*x+6*x^2-4*x^3)/(-1+3*x-6*x^3+4*x^4).
%t Table[1+2^(n+2)-2^(1+n/2)+(-1)^(n+1) 2^(1+n/2)-2^((n+1)/2)+(-1)^(n+2) 2^((n+1)/2), {n,0,28}] (*or*)
%t CoefficientList[Series[(-1 - 2 x + 6 x^2 - 4 x^3)/(-1 + 3 x - 6 x^3 + 4 x^4), {x,0,28}], x] (*or*)
%t LinearRecurrence[{3, 0, -6, 4}, {1, 5, 9, 25}, 29]
%o (PARI) Vec((-1-2*x+6*x^2-4*x^3) / (-1+3*x-6*x^3+4*x^4) + O(x^29))
%Y Cf. A000225, A060867, A092440, A224195.
%K nonn
%O 0,2
%A _Daniel Poveda Parrilla_, Jan 26 2017
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