Search: id:A146362 Results 1-1 of 1 results found. %I A146362 %S A146362 521,617,709,1433,1597,2549,2909,3581,3821,4013,4649,5501, %T A146362 5693,5813,6197,7853,8093,8573,9281,9677,10597,10973,11273,13109,13613, 15413,15641,15737,16001,16477,17093,20261 %N A146362 Primes p such that continued fraction of (1+Sqrt[p])/2 has period 17 : primes in A146340. %t A146362 $MaxExtraPrecision = 4000; s = 10; aa = {}; Do[k = ContinuedFraction[(1 + Sqrt[Prime[n]])/2, 3000]; m = 1; While[k[[s]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]] || k[[s + 4 m]] != k[[s + 5 m]], m++ ]; s = s + 1; While[k[[s]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]] || k[[s + 4 m]] != k[[s + 5 m]], m++ ]; s = s + 1; While[k[[s]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]] || k[[s + 4 m]] != k[[s + 5 m]]; AppendTo[aa, m]], {n, 1, 1495}]; bb = {}; Do[If[aa[[n]] == 17, AppendTo[bb, Prime[n]]], {n, 1, Length[aa]}]; bb (*Artur Jasinski*) %Y A146362 A000290, A050950-A050969, A078370, A146326-A146345, A146348-A146360. %Y A146362 Sequence in context: A094903 A139663 A146340 this_sequence A050966 A113158 A004928 %Y A146362 Adjacent sequences: A146359 A146360 A146361 this_sequence A146363 A146364 A146365 %K A146362 nonn %O A146362 1,1 %A A146362 Artur Jasinski (grafix(AT)csl.pl), Oct 30 2008 %E A146362 Period length in defintion corrected, 2579, 5003 removed, 5813 inserted R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 06 2009 Search completed in 0.001 seconds