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%I #29 May 14 2023 19:39:12
%S 1,4,11,28,68,160,368,832,1856,4096,8960,19456,41984,90112,192512,
%T 409600,868352,1835008,3866624,8126464,17039360,35651584,74448896,
%U 155189248,322961408,671088640,1392508928,2885681152,5972688896,12348030976,25501368320,52613349376
%N a(0) = 1, a(1) = 4, a(2) = 11, and a(n) = 4*a(n-1) - 4*a(n-2) for n >= 3.
%H Peter Kagey, <a href="/A339252/b339252.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4).
%F G.f.: (1 - x^2)/(1 - 2*x)^2.
%F a(n) = A207615(n+2, 2).
%F a(n) = 2^(n-2)*(3*n + 5) for n >= 1. - _Kevin Ryde_, Nov 28 2020
%F E.g.f.: (exp(2*x)*(5 + 6*x) - 1)/4. - _Stefano Spezia_, May 14 2023
%p a := proc(n) option remember; if n <= 2 then return [1, 4, 11][n+1] fi;
%p 4*a(n - 1) - 4*a(n - 2) end: seq(a(n), n = 0..31);
%t CoefficientList[Series[(1 - x^2)/(1 - 2*x)^2, {x, 0, 50}], x]
%Y Cf. A207615, A106472 (quarter).
%K nonn,easy
%O 0,2
%A _Peter Kagey_ and _Peter Luschny_, Nov 28 2020
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