|
Search: id:A071296
|
|
|
| A071296 |
|
a(n) is the least m such that a period of the continued fraction expansion of sqrt(m) is 1,1,1,...,1,1,1,Z and there are n ones in the period (Z is 2*floor(sqrt(m))). If no such m exists, a(n) = 0. |
|
+0
1
|
|
| 3, 0, 7, 13, 0, 58, 135, 0, 819, 2081, 0, 13834, 35955, 0, 244647, 639389, 0, 4374866, 11448871, 0, 78439683, 205337953, 0, 1407271538, 3684200835, 0, 25251313255, 66108441037, 0, 453111560266, 1186259960295, 0
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
FORMULA
|
Let F(n) = n-th Fibonacci number (A000045). If n == 2 mod 3 then F(n+1) is even and there's no such m. Otherwise, let x = (F(n+1) + 1) / 2. Then a(n) = x^2 + [F(n-1) + 2*x*F(n)]/F(n+1).
|
|
EXAMPLE
|
a(3) = 7 because sqrt(7)'s continued fraction is [2;1,1,1,4,...]; the period has 3 ones (and only one other number).
|
|
CROSSREFS
|
Cf. A000045, A010342.
Adjacent sequences: A071293 A071294 A071295 this_sequence A071297 A071298 A071299
Sequence in context: A011200 A021329 A019970 this_sequence A013450 A013421 A011081
|
|
KEYWORD
|
more,nonn
|
|
AUTHOR
|
Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 11 2002
|
|
EXTENSIONS
|
Edited by Don Reble (djr(AT)nk.ca), Jun 06 2003
|
|
|
Search completed in 0.002 seconds
|
