%I #27 Sep 08 2022 08:45:10
%S 1,4,14,47,156,516,1705,5632,18602,61439,202920,670200,2213521,
%T 7310764,24145814,79748207,263390436,869919516,2873148985,9489366472,
%U 31341248402,103513111679,341880583440,1129154862000,3729345169441,12317190370324
%N a(1)=1, a(n) = ceiling(r(3)*a(n-1)) where r(3) = (1/2)*(3 + sqrt(13)) is the positive root of X^2 = 3*X + 1.
%C More generally the sequence a(1)=1, a(n) = ceiling(r(z)*a(n-1)) where r(z) = (1/2)*(z + sqrt(z^2 + 4)) is the positive root of X^2 = z*X + 1 satisfies the linear recurrence: for n > 3, a(n) = (z+1)*a(n-1) - (z-1)*a(n-2) - a(n-3) and the closed-form formula: a(n) = floor(t(z)*r(z)^n) where t(z) = (1/(2*z))*(1+(z+2)/sqrt(z^2+4)) is the positive root of z*(z^2 + 4)*X^2 = (z^2 + 4)*X + 1.
%H Vincenzo Librandi, <a href="/A082574/b082574.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-2,-1).
%F a(1)=1, a(2)=4, a(3)=14, a(n) = 4*a(n-1) - 2*a(n-2) - a(n-3).
%F a(n) = floor(t(3)*r(3)^n) where t(3) = (1/6)*(1 + 5/sqrt(13)) is the positive root of 39*X^2 = 13*X + 1.
%F G.f.: 1/((1-x)*(1-3*x-x^2)). Partial sums of A006190. - _Paul Barry_, Jul 10 2004
%p a:=n->sum(fibonacci(i,3), i=0..n): seq(a(n), n=1..30); # _Zerinvary Lajos_, Mar 20 2008
%t LinearRecurrence[{4, -2, -1}, {1, 4, 14}, 30] (* _Vincenzo Librandi_, Sep 12 2017 *)
%t Table[Sum[Fibonacci[k, 3], {k,0,n}], {n,1,30}] (* _G. C. Greubel_, May 31 2019 *)
%o (PARI) Vec(1/((1-x)*(1-3*x-x^2)) + O(x^30)) \\ _Michel Marcus_, Sep 12 2017
%o (Magma) I:=[1,4,14]; [n le 3 select I[n] else 4*Self(n-1)-2*Self(n-2)-Self(n-3): n in [1..30]]; // _Vincenzo Librandi_, Sep 12 2017
%o (Sage) (1/((1-x)*(1-3*x-x^2))).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, May 31 2019
%Y Cf. A000071, A048739, A049652.
%K nonn
%O 1,2
%A _Benoit Cloitre_, May 06 2003