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%I #53 Jan 25 2024 04:51:28
%S 1,9,89,881,8721,86329,854569,8459361,83739041,828931049,8205571449,
%T 81226783441,804062262961,7959395846169,78789896198729,
%U 779939566141121,7720605765212481,76426118085983689,756540575094624409,7488979632860260401,74133255753507979601,733843577902219535609
%N a(n) = A054320(n) - A001078(n).
%C Numbers k such that 6*k^2 - 2 is a square. - _Bruno Berselli_, Feb 10 2014
%D H. Brocard, Note #2049, L'Intermédiaire des Mathématiciens, 8 (1901), pp. 212-213. - _N. J. A. Sloane_, Mar 02 2022
%H Vincenzo Librandi, <a href="/A138288/b138288.txt">Table of n, a(n) for n = 0..1000</a>
%H Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, <a href="https://arxiv.org/abs/2401.13524">Combinatorics on words and generating Dirichlet series of automatic sequences</a>, arXiv:2401.13524 [math.CO], 2024.
%H Bruno Deschamps, <a href="http://dx.doi.org/10.1016/j.jnt.2010.06.006">Sur les bonnes valeurs initiales de la suite de Lucas-Lehmer</a>, Journal of Number Theory, Volume 130, Issue 12, December 2010, Pages 2658-2670.
%H Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
%H Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_1.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
%H Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_2.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
%H Editors, L'Intermédiaire des Mathématiciens, <a href="/A072256/a072256_3.pdf">Query 4500: The equation x(x+1)/2 = y*(y+1)/3</a>, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10, -1).
%F a(n) = A072256(n+1).
%F a(n) = A001079(n) + 2*A001078(n).
%F a(n) = 10*a(n-1) - a(n-2). a(-1) = a(0) = 1.
%F (sqrt(2)+sqrt(3))^(2*n+1) = A054320(n-1)*sqrt(2) + a(n)*sqrt(3).
%F From _Michael Somos_, Jan 25 2013: (Start)
%F G.f.: (1 - x) / (1 - 10*x + x^2).
%F a(-1-n) = a(n). (End)
%F a(n) = sqrt(2+(5-2*sqrt(6))^(1+2*n)+(5+2*sqrt(6))^(1+2*n))/(2*sqrt(3)). - _Gerry Martens_, Jun 04 2015
%F E.g.f.: exp(5*x)*(3*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/3. - _Stefano Spezia_, May 16 2023
%e 1 + 9*x + 89*x^2 + 881*x^3 + 8721*x^4 + 86329*x^5 + ...
%t CoefficientList[Series[(1 - x)/(1 - 10 x + x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Feb 12 2014 *)
%t a[c_, n_] := Module[{},
%t p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
%t d := Denominator[Convergents[Sqrt[c], n p]];
%t t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
%t Return[t];
%t ] (* Complement of A041007, A041039 *)
%t a[6, 20] (* _Gerry Martens_, Jun 07 2015 *)
%o (Sage) [lucas_number1(n,10,1)-lucas_number1(n-1,10,1) for n in range(1, 20)] # _Zerinvary Lajos_, Nov 10 2009
%o (PARI) {a(n) = subst( poltchebi(n+1) + poltchebi(n), x, 5) / 6} /* _Michael Somos_, Jan 25 2013 */
%Y Cf. A001078, A001079, A072256, A138281.
%Y Cf. similar sequences listed in A238379.
%Y Cf. A041007, A041039, A054320.
%K nonn,easy
%O 0,2
%A _Reinhard Zumkeller_, Mar 12 2008