login
a(1) = 3; for n >= 1, a(2*n) = 2^(n+1), a(2*n+1) = 5*2^(n-1).
0

%I #18 Sep 08 2022 08:46:05

%S 3,4,5,8,10,16,20,32,40,64,80,128,160,256,320,512,640,1024,1280,2048,

%T 2560,4096,5120,8192,10240,16384,20480,32768,40960,65536,81920,131072,

%U 163840,262144,327680,524288,655360,1048576,1310720,2097152,2621440,4194304,5242880

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

%C Union of A020714 and A198633.

%C Essentially the same as A094958.

%C For every n, a(1)^3 + a(2)^3 + a(3)^3 + ... + a(2*n-1)^3 is a cube.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Cube_(algebra)">Cube (algebra)</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,2).

%F a(n) = ceiling((9 - (- 1)^n)*2^(floor(n/2) - 2)).

%F a(n) = n + 2 for n <= 3; a(n) = 2*a(n-2) for n > 3.

%F From _Bruno Berselli_, Aug 20 2013: (Start)

%F G.f.: x*(3+4*x-x^2)/(1-2*x^2).

%F a(n) = (16-(8-5*r)*(1-(-1)^n))*r^(n-6) for n>1, r=sqrt(2). (End)

%e a(9) = 40 because it is equal to 5*2^(4-1).

%t CoefficientList[Series[(3 + 4 x - x^2)/(1 - 2 x^2), {x, 0, 50}], x] (* _Bruno Berselli_, Aug 20 2013 *)

%o (Magma) [n le 3 select n+2 else 2*Self(n-2) : n in [1..43]]

%o (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 Cf. A020714, A094958, A121451, A164682, A198633.

%K nonn,easy

%O 1,1

%A _Arkadiusz Wesolowski_, Aug 20 2013