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A228305 a(1) = 3; for n >= 1, a(2*n) = 2^(n+1), a(2*n+1) = 5*2^(n-1).

%I A228305 #25 Apr 09 2025 18:46:47
%S A228305 3,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048,
%T A228305 2560,4096,5120,8192,10240,16384,20480,32768,40960,65536,81920,131072,
%U A228305 163840,262144,327680,524288,655360,1048576,1310720,2097152,2621440,4194304,5242880
%N A228305 a(1) = 3; for n >= 1, a(2*n) = 2^(n+1), a(2*n+1) = 5*2^(n-1).
%C A228305 Union of A020714 and A198633.
%C A228305 Essentially the same as A094958.
%C A228305 For every n, a(1)^3 + a(2)^3 + a(3)^3 + ... + a(2*n-1)^3 is a cube.
%H A228305 Wikipedia, <a href="http://en.wikipedia.org/wiki/Cube_(algebra)">Cube (algebra)</a>
%H A228305 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,2).
%F A228305 a(n) = ceiling((9 - (- 1)^n)*2^(floor(n/2) - 2)).
%F A228305 a(n) = n + 2 for n <= 3; a(n) = 2*a(n-2) for n > 3.
%F A228305 From _Bruno Berselli_, Aug 20 2013: (Start)
%F A228305 G.f.: x*(3+4*x-x^2)/(1-2*x^2).
%F A228305 a(n) = (16-(8-5*r)*(1-(-1)^n))*r^(n-6) for n>1, r=sqrt(2). (End)
%F A228305 E.g.f.: (8*cosh(sqrt(2)*x) + 5*sqrt(2)*sinh(sqrt(2)*x) + 2*x - 8)/4. - _Stefano Spezia_, Apr 09 2025
%e A228305 a(9) = 40 because it is equal to 5*2^(4-1).
%t A228305 CoefficientList[Series[(3 + 4 x - x^2)/(1 - 2 x^2), {x, 0, 50}], x] (* _Bruno Berselli_, Aug 20 2013 *)
%o A228305 (Magma) [n le 3 select n+2 else 2*Self(n-2) : n in [1..43]];
%o A228305 (PARI) r=43; print1(3); print1(", "); for(n=2, r, if(bitand(n, 1), print1(5*2^((n-3)/2)), print1(2^(n/2+1))); print1(", "));
%Y A228305 Cf. A020714, A094958, A121451, A164682, A198633.
%K A228305 nonn,easy
%O A228305 1,1
%A A228305 _Arkadiusz Wesolowski_, Aug 20 2013