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%I #32 May 08 2024 05:46:54
%S 1,2,4,8,17,38,88,208,497,1194,2876,6936,16737,40398,97520,235424,
%T 568353,1372114,3312564,7997224,19306993,46611190,112529352,271669872,
%U 655869073,1583407994,3822685036,9228778040,22280241089,53789260190,129858761440,313506783040
%N a(n) = Sum_{k=0..floor(n/4)} binomial(n-2k, 2k)*2^(n-4k).
%C Binomial transform of 1,1,1,1,2,2,4,4,8,8,... (g.f.: (1-x)(1+x)^2/(1-2x^2)).
%C Row sums of number triangle A108350. - _Paul Barry_, May 31 2005
%H Vincenzo Librandi, <a href="/A100131/b100131.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4,0,1).
%F G.f.: (1-2x)/((1-2x)^2-x^4) = (1-2x)/((1-x)^2(1-2x-x^2));
%F a(n) = 4a(n-1) - 4a(n-2) + a(n-4);
%F a(n) = ((sqrt(2)+1)^(n+1) + (sqrt(2)-1)^(n+1)(-1)^n)/(4*sqrt(2)) + (n+1)/2;
%F a(n) = Sum_{k=0..n} (1-k)*A000129(n-k+1).
%F a(n) = Sum_{k=0..n} Sum_{j=0..n-k} binomial(k, j)*binomial(n-j, k)*((j+1) mod 2). - _Paul Barry_, May 31 2005
%F a(n) = (1/2)*(Pell(n+1) + n + 1), where Pell(n) = A000129(n). - _Ralf Stephan_, May 15 2007 [corrected by _Jon E. Schoenfield_, Feb 19 2019]
%p with(combinat): seq((n+1+fibonacci(n+1, 2))/2, n=0..30); # _Zerinvary Lajos_, Jun 02 2008
%t CoefficientList[Series[(1-2x)/((1-2x)^2-x^4),{x,0,40}],x] (* _Harvey P. Dale_, Mar 22 2011 *)
%t LinearRecurrence[{4,-4,0,1},{1,2,4,8},40] (* _Vincenzo Librandi_, Jun 25 2012 *)
%o (Magma) I:=[1, 2, 4, 8]; [n le 4 select I[n] else 4*Self(n-1)-4*Self(n-2)+Self(n-4): n in [1..30]]; // _Vincenzo Librandi_, Jun 25 2012
%Y Cf. A098576, A100132, A100133.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Nov 06 2004
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