%I #29 Mar 30 2020 08:36:53

%S 17,37,61,101,197,257,317,401,461,557,577,677,773,1129,1297,1429,1601,

%T 1877,1901,2917,3137,4357,4597,5417,5477,6053,7057,8101,8761,8837,

%U 10733,11621,12101,13457,13877,14401,15277,15377,15877,16333,16901,17737,17957,18329,21317,22501,23593,24337,25601,28901,30137,30977,32401,33857,41453,41617,42437,44101

%N Primes p such that continued fraction of (1 + sqrt(p))/2 has period 3.

%C Primes in A146328. Finite A050952 is subset of this sequence.

%C From _Michel Lagneau_, Sep 03 2014: (Start)

%C 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).

%C 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, ... .

%C (End)

%C 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

%H Amiram Eldar, <a href="/A146348/b146348.txt">Table of n, a(n) for n = 1..2500</a> (terms 1..200 from Zak Seidov)

%p 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

%t okQ[n_] := Length[ContinuedFraction[(1 + Sqrt[n])/2][[2]]] == 3; Select[Prime[Range[100]], okQ]

%Y Cf. A000290, A050952, A078370, A146326-A146345, A146348-A146360, A146363.

%K nonn

%O 1,1

%A _Artur Jasinski_, Oct 30 2008

%E 1019 removed; more terms added by _R. J. Mathar_, Sep 06 2009

%E More terms from _Zak Seidov_, Mar 09 2011