A134410 Second-order Lucas numbers; a(n) = (2n+3)*Lucas(n) - n*Lucas(n-1).
6, 3, 19, 27, 61, 108, 204, 367, 661, 1173, 2069, 3622, 6306, 10923, 18839, 32367, 55421, 94608, 161064, 273527, 463481, 783753, 1322869, 2229002, 3749886, 6299283, 10567579, 17705667, 29630461, 49532148, 82715844, 137997247, 230015581, 383064573, 637434389
Offset: 0
Examples
G.f. = 6 + 3*x + 19*x^2 + 27*x^3 + 61*x^4 + 108*x^5 + 204*x^6 + 367*x^7 + ...
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).
Programs
-
Mathematica
a[ n_] := 3 (n + 1) LucasL[n] - n LucasL[n + 1]; (* Michael Somos, Jun 02 2014 *) LinearRecurrence[{2,1,-2,-1},{6,3,19,27},40] (* Harvey P. Dale, Jun 26 2017 *)
-
PARI
{a(n) = (6 + 5*n) * fibonacci(n+1) - (3 + 5*n) * fibonacci(n)}; /* Michael Somos, Jun 02 2014 */ -
PARI
Vec((2-x)*(3-3*x+2*x^2)/(1-x-x^2)^2 + O(x^40)) \\ Colin Barker, Jun 02 2016
Formula
Defining equation: a(n) = (2n+3)*Lucas(n) - n*Lucas(n-1).
Recurrence: a(0) = 6, a(1) = 3, a(n+2) = a(n+1) + a(n) + 5*Lucas(n).
O.g.f.: (2-x)*(3-3x+2x^2)/(1-x-x^2)^2.
Set A(n) = (a(n-1) + a(n+1))/5, B(n) = a(n+1) - a(n-1). Then A(n+2) = A(n+1) + A(n) + 5*Fibonacci(n) and B(n+2) = B(n+1) + B(n) + 5*Lucas(n). The polynomials L_2(n,-x) = sum {k = 0..n} C(n,k)*a(n-k)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the polynomials L(n,-x) defined in A132148.
0 = a(n)*(-a(n) - 4*a(n+1) + a(n+2)) + a(n+1)*(-3*a(n+1) + 6*a(n+2) - a(n+3)) + a(n+2)*(+3*a(n+2) - 4*a(n+3)) + a(n+3)*(+a(n+3)) for all n in Z. - Michael Somos, Jun 02 2014
a(n) = 2^(-1-n)*(6*((1-sqrt(5))^n+(1+sqrt(5))^n)+(-(-5+sqrt(5))*(1+sqrt(5))^n+(1-sqrt(5))^n*(5+sqrt(5)))*n). - Colin Barker, Jun 02 2016
Comments