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a(n) = Sum_{k=0..floor(n/2)} k*binomial(n,2*k) = floor(n*2^(n-3)).
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%I #32 Feb 14 2023 08:57:52

%S 0,0,1,3,8,20,48,112,256,576,1280,2816,6144,13312,28672,61440,131072,

%T 278528,589824,1245184,2621440,5505024,11534336,24117248,50331648,

%U 104857600,218103808,452984832,939524096,1946157056,4026531840,8321499136,17179869184,35433480192

%N a(n) = Sum_{k=0..floor(n/2)} k*binomial(n,2*k) = floor(n*2^(n-3)).

%C Essentially same as A001792, except for leading zeros, which motivate the existence of this sequence on its own.

%H Vincenzo Librandi, <a href="/A049610/b049610.txt">Table of n, a(n) for n = 0..1000</a>

%H Silvana Ramaj, <a href="https://digitalcommons.georgiasouthern.edu/etd/2273">New Results on Cyclic Compositions and Multicompositions</a>, Master's Thesis, Georgia Southern Univ., 2021. See p. 67.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4).

%F G.f. x^2*(1-x)/(1-2*x)^2. - _Sergei N. Gladkovskii_, Oct 18 2012

%F G.f.: x^2*( 1 + 2*x*U(0) ) where U(k) = 1 + (k+1)/(2 - 8*x/(4*x + (k+1)/U(k+1))); (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Oct 19 2012

%F E.g.f.: x*(exp(2*x) - 1)/4. - _Stefano Spezia_, Feb 02 2023

%F Sum_{n>=2} 1/a(n) = 8*log(2) - 4. - _Amiram Eldar_, Feb 14 2023

%t CoefficientList[Series[x^2*(1 - x)/(1 - 2*x)^2, {x, 0, 40}], x] (* _Vincenzo Librandi_, Jan 09 2013 *)

%o (PARI) a(n)=n<<(n-3)

%Y Cf. A001792, A189162, A189390, A189391.

%K nonn,easy

%O 0,4

%A _M. F. Hasler_, Jan 25 2012