A306672 Partial sums of the even Lucas numbers (A014448).

2, 6, 24, 100, 422, 1786, 7564, 32040, 135722, 574926, 2435424, 10316620, 43701902, 185124226, 784198804, 3321919440, 14071876562, 59609425686, 252509579304, 1069647742900, 4531100550902, 19194049946506, 81307300336924, 344423251294200, 1459000305513722, 6180424473349086

Offset: 0

Author

Rigoberto Florez, Mar 04 2019

Keywords

• nonn

Examples

			L(0) + L(3) = 6;
L(0) + L(3) + L(6) = 24;
L(0) + L(3) + L(6) + L(9) = 100.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..1593
• Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Crossrefs

Cf. A000032, A014448, A099919.

Programs

Maple

f:= gfun:-rectoproc({a(n + 3) - 5*a(n + 2) + 3*a(n + 1) + a(n), a(0) = 2, a(1) = 6, a(2) = 24},a(n),remember):
map(f, [$0..60]); # Robert Israel, Mar 05 2019

Mathematica

Table[(Lucas[3n+2]-1)/2+1,{n,0,25}]
Accumulate[Select[LucasL[Range[0,100]],EvenQ]] (* or *) LinearRecurrence[ {5,-3,-1},{2,6,24},30] (* Harvey P. Dale, Jan 18 2021 *)

PARI

L(n) = fibonacci(n+1)+fibonacci(n-1);
a(n) = sum(k=0, n, L(3*k)); \\ Michel Marcus, Mar 05 2019

Perl

use ntheory ":all"; sub a { vecsum(map{lucasv(1,-1,3*$)}0..$[0]) } # Dana Jacobsen, Mar 05 2019

Formula

a(n) = L(0) + L(3) + L(6) + L(9) + ... + L(3n), L(n) = Lucas numbers A000032.

a(n) = Sum_{i=0..n} L(3i).

a(n) = (L(3*n+2)-1)/2+1.

G.f.: -2*(2*x-1)/((x-1)*(x^2+4*x-1)). - Alois P. Heinz, Mar 04 2019