OFFSET
0,2
COMMENTS
Numbers k such that 6*k^2 - 2 is a square. - Bruno Berselli, Feb 10 2014
REFERENCES
H. Brocard, Note #2049, L'Intermédiaire des Mathématiciens, 8 (1901), pp. 212-213. - N. J. A. Sloane, Mar 02 2022
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, Combinatorics on words and generating Dirichlet series of automatic sequences, arXiv:2401.13524 [math.CO], 2024.
Bruno Deschamps, Sur les bonnes valeurs initiales de la suite de Lucas-Lehmer, Journal of Number Theory, Volume 130, Issue 12, December 2010, Pages 2658-2670.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (10, -1).
FORMULA
a(n) = A072256(n+1).
a(n) = 10*a(n-1) - a(n-2). a(-1) = a(0) = 1.
(sqrt(2)+sqrt(3))^(2*n+1) = A054320(n-1)*sqrt(2) + a(n)*sqrt(3).
From Michael Somos, Jan 25 2013: (Start)
G.f.: (1 - x) / (1 - 10*x + x^2).
a(-1-n) = a(n). (End)
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
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
EXAMPLE
1 + 9*x + 89*x^2 + 881*x^3 + 8721*x^4 + 86329*x^5 + ...
MATHEMATICA
CoefficientList[Series[(1 - x)/(1 - 10 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)
a[c_, n_] := Module[{},
p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
d := Denominator[Convergents[Sqrt[c], n p]];
t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
Return[t];
a[6, 20] (* Gerry Martens, Jun 07 2015 *)
PROG
(Sage) [lucas_number1(n, 10, 1)-lucas_number1(n-1, 10, 1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009
(PARI) {a(n) = subst( poltchebi(n+1) + poltchebi(n), x, 5) / 6} /* Michael Somos, Jan 25 2013 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Reinhard Zumkeller, Mar 12 2008
STATUS
approved