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n * Lucas(n) - (n - 1) * Lucas(n - 1).
1

%I #21 Sep 04 2014 07:24:05

%S 1,5,6,16,27,53,95,173,308,546,959,1675,2909,5029,8658,14852,25395,

%T 43297,73627,124909,211456,357270,602551,1014551,1705657,2863493,

%U 4800990,8039608,13447563,22469261,37505879,62546285,104212364,173489994,288593903,479706787,796815125,1322659237,2194126122,3637574444,6027141411,9980945785

%N n * Lucas(n) - (n - 1) * Lucas(n - 1).

%C By definition, the arithmetic mean of a(1), ... a(n) is equal to L(n) and a(n) - Lucas(n) = (n - 1) * Lucas(n - 2). See A136391 for the Fibonacci case.

%F Recurrence: a(n + 1) = a(n) + a(n - 1) + 5*F(n - 2), n >= 2, where F = A000045. Proof: similar to A136391.

%F Also, a(n) = 2*a(n - 1) + a(n - 2) - 2*a(n - 3) - a(n - 4).

%F G.f.: x*(1 - x)*(1 + 4*x - x^2)/(1 - x - x^2)^2.

%e a(6) = 53 = 6*Lucas(6) - 5*Lucas(5) = 6 * 18 - 5 * 11 = 108 - 55.

%e a(4) = 16 = 4*Lucas(2) + Lucas(3) = 3*Lucas(2) + Lucas(4).

%p with(combinat): seq(n*(fibonacci(n-1)+fibonacci(n-3)) +fibonacci(n)+fibonacci(n-2),n=1..40).

%t Table[LucasL[n]n - LucasL[n - 1](n - 1), {n, 35}] (* _Alonso del Arte_, Sep 02 2014 *)

%o (PARI) a(n) = n*(fibonacci(n-1)+fibonacci(n-3)) +fibonacci(n)+fibonacci(n-2); \\ _Michel Marcus_, Sep 02 2014

%Y Cf. A000032, A000045, A136391.

%K nonn,easy

%O 1,2

%A _Giuseppe Coppoletta_, Sep 02 2014