OFFSET
0,2
COMMENTS
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 16's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,15}. - Milan Janjic, Jan 23 2015
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..800
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (16,-1).
FORMULA
a(n) = 16*a(n-1) - a(n-2), n>=1, a(-1)=0, a(0)=1.
a(n) = S(n, 16) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. See A049310.
G.f.: 1/(1 - 16*x + x^2).
a(n) = (((8 + 3*sqrt(7))^(n+1) - (8 - 3*sqrt(7))^(n+1)))/(6*sqrt(7)).
a(n) = sqrt(A001081(n+1)^2-1)/63).
a(n) = Sum_{k=0..n} A101950(n,k)*15^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/7*(7 + 3*sqrt(7)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/16*(7 + 3*sqrt(7)). - Peter Bala, Dec 23 2012
MATHEMATICA
lst={}; Do[AppendTo[lst, GegenbauerC[n, 1, 8]], {n, 0, 8^2}]; lst (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[1/(1 - 16 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 24 2012 *)
LinearRecurrence[{16, -1}, {1, 16}, 30] (* G. C. Greubel, Jan 18 2018 *)
PROG
(Sage) [lucas_number1(n, 16, 1) for n in xrange(1, 20)] # Zerinvary Lajos, Jun 25 2008
(MAGMA) I:=[1, 16, 255]; [n le 3 select I[n] else 16*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012
(PARI) x='x+O('x^30); Vec(1/(1 - 16*x + x^2)) \\ G. C. Greubel, Jan 18 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Nov 08 2002
STATUS
reviewed