OFFSET
0,3
COMMENTS
The ratio a(n+1)/a(n) converges to (1+sqrt(65))/2 as n approaches infinity. - Felix P. Muga II, Mar 12 2014
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
J. Borowska, L. Lacinska, Recurrence form of determinant of a heptadiagonal symmetric Toeplitz matrix, J. Appl. Math. Comp. Mech. 13 (2014) 19-16, remark 2 for permanent of tridiagonal Toeplitz matrices a=1, b=4.
Index entries for linear recurrences with constant coefficients, signature (1,16).
FORMULA
a(0)=a(1)=1, a(n) = a(n-1) + 16*a(n-2) for n>1.
a(n) = (1/sqrt(65))*( ((1+sqrt(65))/2)^(n+1) - ((1-sqrt(65))/2)^(n+1) ), for n >= 0 [Binet representation] - Felix P. Muga II, Mar 12 2014
E.g.f.: (1/sqrt(65))*exp(x/2)*( sinh((sqrt(65)/2)*x) + sqrt(65)*cosh((sqrt(65)/2) *x) ). - G. C. Greubel, Jul 26 2016
MATHEMATICA
Join[{a=1, b=1}, Table[c=1*b+16*a; a=b; b=c, {n, 60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 01 2011 *)
CoefficientList[Series[1/(1 - x - 16 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 14 2014 *)
LinearRecurrence[{1, 16}, {1, 1}, 30] (* Harvey P. Dale, Aug 14 2014 *)
PROG
(PARI) a(n)=([0, 1; 16, 1]^n*[1; 1])[1, 1] \\ Charles R Greathouse IV, Jul 26 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Philippe Deléham, Nov 30 2009
STATUS
approved