Search: id:A080039
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%I A080039
%S A080039 1,2,5,14,33,82,197,478,1153,2786,6725,16238,39201,94642,228485,551614,
%T A080039 1331713,3215042,7761797,18738638,45239073,109216786,263672645,
%U A080039 636562078,1536796801,3710155682,8957108165,21624372014,52205852193
%N A080039 a(n)=floor((1+sqrt(2))^n).
%C A080039 a(n)=P(n)-(1+(-1)^n)/2, where P(n) is the Pell sequence (A000129) with 
               initial conditions 2, 2.
%C A080039 For n>0 a(n) is the maximum element in the continued fraction for P(n)*sqrt(2) 
               where P=A000129 - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 
               19 2005
%F A080039 G.f.: g(t)=(1-t^2+2t^3)/(1-2t-2t^2+2t^3+t^4)
%F A080039 Comments from Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jan 
               02 2009 (Start): The fractional part of (1+sqrt(2))^n equals (1+sqrt(2))^(-n), 
               if n odd. For even n, the fractional part of (1+sqrt(2))^n is equal 
               to 1-(1+sqrt(2))^(-n).
%F A080039 fract((1+sqrt(2))^n)) = (1/2)*(1+(-1)^n)-(-1)^n*(1+sqrt(2))^(-n) = (1/
               2)*(1+(-1)^n)-(1-sqrt(2))^n.
%F A080039 See A001622 for a general formula concerning the fractional parts of 
               powers of numbers x>1, which suffice x-x^(-1)=floor(x).
%F A080039 a(n) = (sqrt(2)+1)^n - (1/2) + (-1)^n*((sqrt(2)-1)^n - (1/2)) for n>0. 
               (End)
%t A080039 CoefficientList[Series[(1-t^2+2t^3)/(1-2t-2t^2+2t^3+t^4), {t, 0, 30}], 
               t]
%Y A080039 Cf. A001622, A006497, A014176, A098316.
%Y A080039 Sequence in context: A096772 A090803 A018015 this_sequence A131408 A137917 
               A102714
%Y A080039 Adjacent sequences: A080036 A080037 A080038 this_sequence A080040 A080041 
               A080042
%K A080039 easy,nonn
%O A080039 0,2
%A A080039 Mario Catalani (mario.catalani(AT)unito.it), Jan 21 2003

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