OFFSET
1,1
COMMENTS
From Michel Lagneau, Sep 03 2014: (Start)
The primes of the form p = n^2+1 for n>2 are in the sequence, and the continued fraction of (1+sqrt(p))/2 is [n/2; 1, 1, n-1, 1, 1, n-1, 1, 1, ...] with the period (1, 1, n-1).
We observe that the other primes {61, 317, 461, 557, 773, 1129, 1429, ...} are prime divisors of composites numbers of the form k^2+1 where k = 11, 114, 48, 118, 317, 168, 620, ... .
(End)
Another possibly infinite subset of the sequence is primes of the form 100*k^2-44*k+5, where the continued fraction is [5*k-1; 2, 2, 10*k-3, ...] with period [2, 2, 10*k-3]. This includes {61, 317, 773, 1429, 4597, 6053, ...}. - Robert Israel, Sep 03 2014
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..2500 (terms 1..200 from Zak Seidov)
MAPLE
A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic', 'quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146348 := proc(n) RETURN(isprime(n) and A146326(n) = 3) ; end: for n from 2 to 4000 do if isA146348(n) then printf("%d, \n", n) ; fi; od: # R. J. Mathar, Sep 06 2009
MATHEMATICA
okQ[n_] := Length[ContinuedFraction[(1 + Sqrt[n])/2][[2]]] == 3; Select[Prime[Range[100]], okQ]
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 30 2008
EXTENSIONS
1019 removed; more terms added by R. J. Mathar, Sep 06 2009
More terms from Zak Seidov, Mar 09 2011
STATUS
approved