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%I #15 Jun 26 2017 17:57:57
%S 6,3,19,27,61,108,204,367,661,1173,2069,3622,6306,10923,18839,32367,
%T 55421,94608,161064,273527,463481,783753,1322869,2229002,3749886,
%U 6299283,10567579,17705667,29630461,49532148,82715844,137997247,230015581,383064573,637434389
%N Second-order Lucas numbers; a(n) = (2n+3)*Lucas(n) - n*Lucas(n-1).
%C This sequence is defined by analogy with the sequence of second-order Fibonacci numbers A010049.
%C Inverse binomial transform is (-1)^n * a(n). - _Michael Somos_, Jun 02 2014
%H Colin Barker, <a href="/A134410/b134410.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-2,-1).
%F Defining equation: a(n) = (2n+3)*Lucas(n) - n*Lucas(n-1).
%F Recurrence: a(0) = 6, a(1) = 3, a(n+2) = a(n+1) + a(n) + 5*Lucas(n).
%F O.g.f.: (2-x)*(3-3x+2x^2)/(1-x-x^2)^2.
%F 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.
%F 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
%F 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
%e G.f. = 6 + 3*x + 19*x^2 + 27*x^3 + 61*x^4 + 108*x^5 + 204*x^6 + 367*x^7 + ...
%t a[ n_] := 3 (n + 1) LucasL[n] - n LucasL[n + 1]; (* _Michael Somos_, Jun 02 2014 *)
%t LinearRecurrence[{2,1,-2,-1},{6,3,19,27},40] (* _Harvey P. Dale_, Jun 26 2017 *)
%o (PARI) {a(n) = (6 + 5*n) * fibonacci(n+1) - (3 + 5*n) * fibonacci(n)}; /* _Michael Somos_, Jun 02 2014 */
%o (PARI) Vec((2-x)*(3-3*x+2*x^2)/(1-x-x^2)^2 + O(x^40)) \\ _Colin Barker_, Jun 02 2016
%Y Cf. A000032, A010049, A132148.
%K easy,nonn
%O 0,1
%A _Peter Bala_, Oct 24 2007
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