OFFSET
0,2
COMMENTS
a(n)^2 - 42 (2*b(n))^2 = +1 with b(n):=A097309(n) gives all nonnegative solutions of this D:= 42*4 = 168 Pell equation.
Numbers n such that 42*(n^2-1) is a square. - Vincenzo Librandi, Nov 17 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 26xy + y^2 + 168 = 0. - Colin Barker, Feb 20 2014
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..700 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (26,-1).
FORMULA
a(n) = 26*a(n-1) - a(n-2), a(-1) := 13, a(0)=1.
a(n) = T(n, 13) = (S(n, 26) - S(n-2, 26))/2 = S(n, 26) - 13*S(n-1, 26) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(n, 26)=A097309(n).
a(n) = (ap^n + am^n)/2 with ap := 13+2*sqrt(42) and am := 13-2*sqrt(42).
a(n) = Sum_{k=0..floor(n/2)} ((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*13)^(n-2*k), n >= 1.
G.f.: (1 - 13*x)/(1 - 26*x + x^2).
a(n) = sqrt(1 + 168*A097309(n)^2), n >= 0.
a(n) = cosh(2n*arcsinh(sqrt(6))). - Herbert Kociemba, Apr 24 2008
MATHEMATICA
CoefficientList[Series[(1 - 13 x)/(1 - 26 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 22 2014 *)
LinearRecurrence[{26, -1}, {1, 13}, 20] (* Harvey P. Dale, Jul 01 2019 *)
PROG
(PARI) Vec((1-13*x)/(1-26*x+x^2) + O(x^100)) \\ Colin Barker, Feb 20 2014
(Magma) I:=[1, 13]; [n le 2 select I[n] else 26*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 22 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 31 2004
EXTENSIONS
Additional terms from Colin Barker, Feb 20 2014
STATUS
approved