%I A156845
%S A156845 3588,15755,27922,40089,52256,64423,76590,88757,100924,113091,125258,
%T A156845 137425,149592,161759,173926,186093,198260,210427,222594,234761,246928,
%U A156845 259095,271262,283429,295596,307763,319930,332097,344264,356431,368598
%N A156845 a(n)=12167*n-8579 (n>0)
%C A156845 Arises in solving Pell equations of the form X^2 - A*Y^2 = 1.
%C A156845 Let n=[A156849] (156,373,685,902,...,) =n^2-2=0 mod (23^2). If A=[A156841] 
               (46,263,1538,3871,.,) = 529*n^2-312*n+46 or A=[156842] (263,46,887,
               2787) =(529*n^2-746*n+263 , Y=23*n, or [A156845] (3588,15755,27922,
               ...,) = 12167*n-8579, or Y=[A156846] (8579,20746,32913,...,) =12167*n-3588, 
               and X=279841*n^2-165048*n+24335 [A156843] (24335,139128,813603,...,
               ) or X=[A156844] =279841*n^2-394634*n+139128 (139128,24335,469224,
               1473795,...,) , we have for all terms, Pell's equation X^2-A*Y^2=1. 
               Example: For n=156, A=46, Y=3588, X=24335, 24335^2-46*3588^2=1 ; 
               n=373, A=263, Y=8579, X=139128; 139128^2-263*8579^2=1; n=685, A=887, 
               Y=15755, X=469224; 469224^2-887*15755^2=1; n=902, A=1538, Y=20746, 
               X=813603; 813603^2-1538*20746^2=1. [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), 
               Feb 20 2009]
%H A156845 Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5773864&tstart=0">
               X^2-AY^2=1</a> [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), 
               Feb 20 2009]
%e A156845 For n=1, a(1)=3588; n=2, a(2)=15755; n=3, a(3)=27922
%Y A156845 Cf. A156846
%Y A156845 Cf. A156849, A156844, A156843, A156842, A156841 [From Vincenzo Librandi 
               (vincenzo.librandi(AT)tin.it), Feb 20 2009]
%Y A156845 Sequence in context: A004932 A004952 A004972 this_sequence A157857 A141781 
               A096472
%Y A156845 Adjacent sequences: A156842 A156843 A156844 this_sequence A156846 A156847 
               A156848
%K A156845 nonn
%O A156845 1,1
%A A156845 Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Feb 17 2009