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a(n)=a(n-1)+a(n-2)+a(n-3)+2a(n-4) with a(n)=n+1 for n<=3.
4

%I #17 Dec 12 2023 08:35:54

%S 1,2,3,4,11,22,43,84,171,342,683,1364,2731,5462,10923,21844,43691,

%T 87382,174763,349524,699051,1398102,2796203,5592404,11184811,22369622

%N a(n)=a(n-1)+a(n-2)+a(n-3)+2a(n-4) with a(n)=n+1 for n<=3.

%H Harvey P. Dale, <a href="/A139763/b139763.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,1,2)

%F a(n+1)-2a(n)=period 4:repeat 0, -1, -2, 3.

%F O.g.f.: (x-1)*(2*x^2+2*x+1)/((2*x-1)(1+x)(x^2+1)). a(n) = A056594(n+3)+((-1)^n+2^(n+1))/3. - _R. J. Mathar_, May 22 2008

%t LinearRecurrence[{1,1,1,2},{1,2,3,4},30] (* _Harvey P. Dale_, May 22 2021 *)

%K nonn

%O 0,2

%A _Paul Curtz_, May 20 2008

%E More terms from _R. J. Mathar_, May 22 2008