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a(n) = n*2^floor((n+1)/2).
3

%I #25 Aug 13 2023 00:51:32

%S 0,2,4,12,16,40,48,112,128,288,320,704,768,1664,1792,3840,4096,8704,

%T 9216,19456,20480,43008,45056,94208,98304,204800,212992,442368,458752,

%U 950272,983040,2031616,2097152,4325376,4456448,9175040,9437184,19398656,19922944,40894464

%N a(n) = n*2^floor((n+1)/2).

%H G. C. Greubel, <a href="/A132314/b132314.txt">Table of n, a(n) for n = 0..1000</a>

%H Simon Plouffe, <a href="http://plouffe.fr/Simon%20Plouffe_fichiers/pibase2integerpart.gif">Illustration</a>.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,4,0,-4).

%F a(n) = 2*A093968(n).

%F From _Chai Wah Wu_, May 30 2016: (Start)

%F a(n) = 4*a(n-2) - 4*a(n-4) for n > 3.

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

%F E.g.f.: x*(2*cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x)). - _G. C. Greubel_, May 30 2016

%F Sum_{n>=1} 1/a(n) = log(2)/2 + log(1+sqrt(2))/sqrt(2). - _Amiram Eldar_, Feb 13 2023

%p seq(n*2^(floor((n+1)/2)),n=0..120);

%t LinearRecurrence[{0,4,0,-4}, {0, 2, 4, 12}, 50] (* _G. C. Greubel_, May 30 2016 *)

%Y Cf. A093968, A075554.

%K nonn,easy

%O 0,2

%A _Simon Plouffe_, Nov 19 2007