OFFSET
0,1
COMMENTS
From Johannes W. Meijer, Jun 12 2010: (Start)
The a(n) terms of this sequence can be constructed with the terms of sequence A090301.
For the terms of the periodical sequence of the continued fraction for sqrt(229) see A040213. We observe that its period is five. (End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 3420, 0, 0, 0, 0, 1).
FORMULA
From Johannes W. Meijer, Jun 12 2010: (Start)
a(5n) = A090301(3n+1), a(5n+1) = (A090301(3n+2) - A090301(3n+1))/2, a(5n+2) = (A090301(3n+2) + A090301(3n+1))/2, a(5n+3) = A090301(3n+2) and a(5n+4) = A090301(3n+3)/2. (End)
G.f.: -(x^9-15*x^8+106*x^7-121*x^6+227*x^5+1710*x^4+227*x^3+121*x^2+106*x+15) / (x^10+3420*x^5-1). - Colin Barker, Nov 08 2013
MATHEMATICA
Numerator[Convergents[Sqrt[229], 30]] (* Vincenzo Librandi, Nov 01 2013 *)
LinearRecurrence[{0, 0, 0, 0, 3420, 0, 0, 0, 0, 1}, {15, 106, 121, 227, 1710, 51527, 362399, 413926, 776325, 5848201}, 30] (* Harvey P. Dale, Dec 19 2016 *)
CROSSREFS
KEYWORD
nonn,frac,cofr,easy
AUTHOR
EXTENSIONS
More terms from Colin Barker, Nov 08 2013
STATUS
approved