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A307268 Partial sums of the Lucas numbers of the form L(3n+2) (A163063).

%I A307268 #47 Aug 10 2022 20:03:43
%S A307268 3,14,61,260,1103,4674,19801,83880,355323,1505174,6376021,27009260,
%T A307268 114413063,484661514,2053059121,8696898000,36840651123,156059502494,
%U A307268 661078661101,2800374146900,11862575248703,50250675141714,212865275815561,901711778403960,3819712389431403
%N A307268 Partial sums of the Lucas numbers of the form L(3n+2) (A163063).
%H A307268 Colin Barker, <a href="/A307268/b307268.txt">Table of n, a(n) for n = 0..1000</a>
%H A307268 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-3,-1).
%F A307268 a(n) = A001076(n+1) + A099919(n+1).
%F A307268 a(n) = Sum_{i=0..n} L(3i+2), L(i) = A000032(i).
%F A307268 a(n) = (L(3*n+4)-1)/2.
%F A307268 From _Colin Barker_, Apr 02 2019: (Start)
%F A307268 G.f.: (3 - x) / ((1 - x)*(1 - 4*x - x^2)).
%F A307268 a(n) = (-2 + (7-3*sqrt(5))*(2-sqrt(5))^n + (2+sqrt(5))^n*(7+3*sqrt(5))) / 4.
%F A307268 a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3) for n > 2.
%F A307268 (End)
%e A307268 L(2) + L(5) = 14;
%e A307268 L(2) + L(5) + L(8) = 61;
%e A307268 L(2) + L(5) + L(8) + L(11) = 260.
%t A307268 Table[(LucasL[3*n + 4] - 1)/2, {n, 0, 20}]
%t A307268 LinearRecurrence[{5,-3,-1},{3,14,61},30] (* _Harvey P. Dale_, Aug 10 2022 *)
%o A307268 (PARI) L(n) = fibonacci(n+1)+fibonacci(n-1);
%o A307268 a(n) = (L(3*n+4)-1)/2; \\ _Michel Marcus_, Apr 01 2019
%o A307268 (PARI) Vec((3 - x) / ((1 - x)*(1 - 4*x - x^2)) + O(x^25)) \\ _Colin Barker_, Apr 02 2019
%Y A307268 Cf. A001076, A014448, A049651, A099919, A163063.
%K A307268 nonn,easy
%O A307268 0,1
%A A307268 _Rigoberto Florez_, Apr 01 2019