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%I #31 Sep 12 2021 18:01:15
%S 3,5,10,18,36,68,136,264,528,1040,2080,4128,8256,16448,32896,65664,
%T 131328,262400,524800,1049088,2098176,4195328,8390656,16779264,
%U 33558528,67112960,134225920,268443648
%N a(n) = (2^n + 2^[ n/2 ] )/2.
%C a(n) is union of A007582(n-1) and A164051(n). - _Jaroslav Krizek_, Aug 14 2009
%H James Spahlinger, <a href="/A001445/b001445.txt">Table of n, a(n) for n = 2..1001</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-4).
%F a(n) = (1/2)*A005418(n+2).
%F G.f.: x^2*(3-x-6*x^2)/((1-2*x)*(1-2*x^2)).
%F G.f.: 3*G(0) where G(k) = 1 + x*(4*2^k + 1)*(1 + 2*x*G(k+1))/(1 + 2*2^k). - _Sergei N. Gladkovskii_, Dec 12 2011 [Edited by _Michael Somos_, Sep 09 2013]
%F a(n) = 2*a(n-1) + 2*a(n-2) - 4*a(n-3) for n > 4. - _Chai Wah Wu_, Sep 10 2020
%e G.f. = 3*x^2 + 5*x^3 + 10*x^4 + 18*x^5 + 36*x^6 + 68*x^7 + 136*x^8 + ...
%p f := n->(2^n+2^floor(n/2))/2;
%t Table[(2^n + 2^(Floor[n/2]))/2, {n, 2, 50}] (* _G. C. Greubel_, Sep 08 2017 *)
%t LinearRecurrence[{2,2,-4},{3,5,10},30] (* _Harvey P. Dale_, Sep 12 2021 *)
%o (PARI) for(n=2,50, print1((2^n + 2^(n\2))/2, ", ")) \\ _G. C. Greubel_, Sep 08 2017
%Y Cf. A056309, A052957.
%K nonn
%O 2,1
%A _N. J. A. Sloane_
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