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Primes p such that continued fraction of (1 + sqrt(p))/2 has period 15: primes in A146338.
37

%I #16 Mar 30 2020 08:42:34

%S 193,281,1861,1933,2089,2141,2437,2741,2837,3037,3121,3413,4001,4637,

%T 4877,5821,6653,7673,8117,10069,10273,10457,11197,11549,11821,12409,

%U 13037,14653,15061,15077,18661,20549,22921,23117,24169,25621,28837,35597,35869,36389,38569

%N Primes p such that continued fraction of (1 + sqrt(p))/2 has period 15: primes in A146338.

%H Amiram Eldar, <a href="/A146360/b146360.txt">Table of n, a(n) for n = 1..2000</a>

%p A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146360 := proc(n) RETURN(isprime(n) and A146326(n) = 15) ; end: for n from 2 to 30000 do if isA146360(n) then printf("%d,\n",n) ; fi; od: # _R. J. Mathar_, Sep 06 2009

%t Select[Prime[Range[1500]],Length[ContinuedFraction[(Sqrt[#]+1)/2][[2]]] == 15&] (* _Harvey P. Dale_, Aug 16 2014 *)

%Y Cf. A000290, A050950-A050969, A078370, A146326-A146345, A146348-A146360.

%K nonn

%O 1,1

%A _Artur Jasinski_, Oct 30 2008

%E 8539 removed by _R. J. Mathar_, Sep 06 2009

%E More terms from _Amiram Eldar_, Mar 30 2020