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A085696 a(n) = L(n) * L(n+1) * L(n+2) / 2, where L(n) = Lucas number (A000032).

%I A085696 #25 Jun 07 2022 10:42:50
%S A085696 3,6,42,154,693,2871,12267,51794,219678,930126,3940797,16692319,
%T A085696 70711683,299536446,1268861682,5374976354,22768778133,96450071031,
%U A085696 408569091147,1730726388874,7331474722278,31056625155606,131557975542717,557288527006079,2360712084085443
%N A085696 a(n) = L(n) * L(n+1) * L(n+2) / 2, where L(n) = Lucas number (A000032).
%H A085696 Michael De Vlieger, <a href="/A085696/b085696.txt">Table of n, a(n) for n = 0..1594</a>
%H A085696 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,6,-3,-1).
%F A085696 G.f.: ( 3 - 3 x + 6 x^2 + x^3 )/( 1 - 3 x - 6 x^2 + 3 x^3 + x^4 ).
%F A085696 Recurrence: a(n+4) = 3 a(n+3) + 6 a(n+2) - 3 a(n+1) - a(n).
%F A085696 From _Michael Somos_, Sep 17 2014: (Start)
%F A085696 a(n) = (6*F(n+1)^3 + 11*F(n+1)^2*F(n) - 3*F(n+1)*F(n)^2 - 2*F(n)^3) / 2 for all n in Z, where F(n) = Fibonacci number (A000045).
%F A085696 a(n) = (2*F(n+1) - F(n)) * (3*F(n+1) + F(n)) * (F(n+1) + 2*F(n)) / 2 for all n in Z.
%F A085696 a(-2-n) = -(-1)^n * a(n) for all n in Z.
%F A085696 0 = a(n)*(-a(n+1) - a(n+2)) + a(n+1)*(-3*a(n+1) + a(n+2)) for all n in Z. (End)
%F A085696 a(n) = -1 + Sum_{k=0..n} L(k)^2 * L(k+1). - _Amiram Eldar_, Jan 13 2022
%e A085696 G.f. = 3 + 6*x + 42*x^2 + 154*x^3 + 693*x^4 + 2871*x^5 + 12267*x^6 + ...
%t A085696 Array[(Times @@ Map[LucasL, # + {0, 1, 2}])/2 &, 25, 0] (* _Michael De Vlieger_, Mar 19 2021 *)
%t A085696 Times@@#/2&/@Partition[LucasL[Range[0,30]],3,1] (* _Harvey P. Dale_, Jun 07 2022 *)
%o A085696 (PARI) {a(n) = my(s); n++; s=if(n<0, -1, 1); n=abs(n); s^n * polcoeff( (-1 + 6*x +3*x^2 + 3*x^3) / ((1 + x - x^2) * (1 - 4*x - x^2)) + x * O(x^n), n)}; /* _Michael Somos_, Sep 17 2014 */
%Y A085696 Cf. A000032, A000045.
%K A085696 easy,nonn
%O A085696 0,1
%A A085696 _Emanuele Munarini_, Jul 18 2003