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A015520 a(n) = 2*a(n-1) + 11*a(n-2), a(0) = 0, a(1) = 1.

%I A015520 #42 Sep 08 2022 08:44:40
%S A015520 0,1,2,15,52,269,1110,5179,22568,102105,452458,2028071,9033180,
%T A015520 40375141,180115262,804357075,3589982032,16027891889,71545586130,
%U A015520 319397983039,1425797413508,6364972640445,28413716829478
%N A015520 a(n) = 2*a(n-1) + 11*a(n-2), a(0) = 0, a(1) = 1.
%H A015520 G. C. Greubel, <a href="/A015520/b015520.txt">Table of n, a(n) for n = 0..1000</a> (terms 0..100 from Vincenzo Librandi)
%H A015520 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2, 11).
%F A015520 a(n) = 2*a(n-1) + 11*a(n-2).
%F A015520 E.g.f.: exp(x)*sinh(2*sqrt(3)*x)/(2*sqrt(3)). - _Paul Barry_, Nov 20 2003
%F A015520 a(n) = Sum_{k=0..n} binomial(n, 2*k+1)*12^k. - _Paul Barry_, Sep 29 2004
%F A015520 From _R. J. Mathar_, Apr 29 2008: (Start)
%F A015520 O.g.f.: x/(1 - 2*x - 11*x^2).
%F A015520 a(n) = 11^n*(A^n - B^n)/(4*sqrt(3)) where A = 1/(2*sqrt(3)-1) and B  = -1/(2*sqrt(3)+1). (End)
%F A015520 a(n) = (Ap^n - Am^n)/(Ap - Am), where Ap = 1 + 2*sqrt(3) and Am = 1 - 2*sqrt(3). (Binet - de Moivre type formula.) This coincides with the preceding formula. - _Wolfdieter Lang_, Feb 17 2018
%t A015520 Join[{a=0,b=1},Table[c=2*b+11*a;a=b;b=c,{n,100}]] (* _Vladimir Joseph Stephan Orlovsky_, Jan 16 2011 *)
%t A015520 LinearRecurrence[{2,11},{0,1},30] (* _Harvey P. Dale_, Jul 13 2011 *)
%o A015520 (Sage) [lucas_number1(n,2,-11) for n in range(0, 23)] # _Zerinvary Lajos_, Apr 22 2009
%o A015520 (Magma) [ n eq 1 select 0 else n eq 2 select 1 else 2*Self(n-1)+11*Self(n-2): n in [1..30] ]; // _Vincenzo Librandi_, Aug 23 2011
%o A015520 (PARI) x='x+O('x^30); concat([0], Vec(-x/(-1+2*x+11*x^2))) \\ _G. C. Greubel_, Jan 01 2018
%Y A015520 Cf. A090042.
%K A015520 nonn,easy
%O A015520 0,3
%A A015520 _Olivier Gérard_