OFFSET
0,2
COMMENTS
Numbers n such that 30*(n^2-1) is square. - Vincenzo Librandi, Aug 08 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 22xy + y^2 + 120 = 0. - Colin Barker, Feb 19 2014
LINKS
Colin Barker, Table of n, a(n) for n = 0..745
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (22,-1).
FORMULA
a(n+1)^2 - 30*(2*b(n))^2 = 1, n>=0, with the companion sequence b(n)=A077421(n).
a(n) = 22*a(n-1) - a(n-2), a(-1) := 11, a(0)=1.
a(n) = T(n, 11) = (S(n, 22)-S(n-2, 22))/2 = S(n, 22)-11*S(n-1, 22) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(n, 22)=A077421(n).
a(n) = (ap^n + am^n)/2 with ap := 11+2*sqrt(30) and am := 11-2*sqrt(30).
a(n) = sum(((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*11)^(n-2*k), k=0..floor(n/2)), n>=1.
a(n+1) = sqrt(1 + 30*(2*A077421(n))^2), n>=0.
a(n) = Cosh[2n*ArcSinh[Sqrt[5]]] - Herbert Kociemba, Apr 24 2008
G.f.: (1-11*x)/(1-22*x+x^2). - Philippe Deléham, Nov 17 2008
MATHEMATICA
Table[Cos[n*ArcCos[11]] // Round, {n, 0, 15}] (* Jean-François Alcover, Dec 19 2013 *)
LinearRecurrence[{22, -1}, {1, 11}, 20] (* Harvey P. Dale, Jul 30 2022 *)
PROG
(Sage) [lucas_number2(n, 22, 1)/2 for n in range(0, 20)] # Zerinvary Lajos, Jun 26 2008
(Magma) [n: n in [1..10000000] |IsSquare(30*(n^2-1))] // Vincenzo Librandi, Aug 08 2010
(PARI) Vec((1-11*x)/(1-22*x+x^2) + O(x^100)) \\ Colin Barker, Jun 15 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Nov 29 2002
STATUS
approved