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a(n) = 2*a(n-1) + 2*a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.
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%I #32 Mar 11 2024 08:30:26

%S 1,4,9,20,49,120,289,696,1681,4060,9801,23660,57121,137904,332929,

%T 803760,1940449,4684660,11309769,27304196,65918161,159140520,

%U 384199201,927538920,2239277041,5406093004,13051463049,31509019100,76069501249

%N a(n) = 2*a(n-1) + 2*a(n-3) + a(n-4), a(0) = 1, a(1) = 4, a(2) = 9, a(3) = 20.

%C Let (b(n)) be the p-INVERT of (1,2,2,2,2,2,...) using p(S) = 1 - S^2; then

%C b(0) = 0 and b(n) = a(n-1) for n >= 1; see A292400. - _Clark Kimberling_, Sep 30 2017

%C Floretion Algebra Multiplication Program, FAMP Code: 2kbasekseq[J+G] with J = + j' + k' + 'ii' and G = + .5'ii' + .5'jj' + .5'kk' + .5e

%H Vincenzo Librandi, <a href="/A111587/b111587.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 (2,0,2,1).

%F a(2n) = A090390(n+1), a(2n+1) = A046729(n+1);

%F G.f.: (x+1)^2/((x^2+1)*(1-2*x-x^2)). [sign flipped by _R. J. Mathar_, Nov 10 2009]

%F a(n) = A057077(n+1)/2 - A001333(n+2)/2. - _R. J. Mathar_, Nov 10 2009

%t LinearRecurrence[{2,0,2,1},{1,4,9,20},30] (* _Harvey P. Dale_, Jul 26 2011 *)

%t CoefficientList[Series[(x + 1)^2 / ((x^2 + 1) (1 - 2 x - x^2)), {x, 0, 33}], x] (* _Vincenzo Librandi_, Oct 01 2017 *)

%o (Magma) I:=[1,4,9,20]; [n le 4 select I[n] else 2*Self(n-1) +2*Self(n-3)+Self(n-4): n in [1..35]]; // _Vincenzo Librandi_, Oct 01 2017

%Y Cf. A019898, A046729, A090390, A111588.

%K easy,nonn

%O 0,2

%A _Creighton Dement_, Aug 08 2005