login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A146327 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 2. 3
2, 3, 10, 11, 12, 15, 21, 26, 27, 30, 35, 45, 50, 51, 56, 63, 77, 82, 83, 84, 87, 90, 93, 99, 117, 122, 123, 132, 143, 165, 170, 171, 182, 195, 221, 226, 227, 228, 230, 231, 235, 237, 240, 245, 255, 285, 290, 291, 306, 323, 357, 362, 363, 380, 399, 437, 442, 443 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
For primes in this sequence see A056899, primes of the form k^2 + 2.
LINKS
EXAMPLE
a(1) = 2 because continued fraction of (1 + sqrt(2))/2 = 1, 4, 1, 4, 1, 4, 1, ... has repeating part (1,4), period 2.
MAPLE
A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146327 := proc(n) RETURN(A146326(n) = 2) ; end: for n from 2 to 450 do if isA146327(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 06 2009
MATHEMATICA
Select[Range[1000], 2 == Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] &]
CROSSREFS
Sequence in context: A007961 A212067 A060811 * A344939 A278742 A250174
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 30 2008
EXTENSIONS
226, 227, 290, 291 added by R. J. Mathar, Sep 06 2009
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified March 28 09:04 EDT 2024. Contains 371240 sequences. (Running on oeis4.)