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 A080039 a(n) = floor((1+sqrt(2))^n). 10

%I

%S 1,2,5,14,33,82,197,478,1153,2786,6725,16238,39201,94642,228485,

%T 551614,1331713,3215042,7761797,18738638,45239073,109216786,263672645,

%U 636562078,1536796801,3710155682,8957108165,21624372014,52205852193

%N a(n) = floor((1+sqrt(2))^n).

%C a(n) = P(n) - (1+(-1)^n)/2, where P(n) is the Pell sequence (A000129) with initial conditions 2, 2.

%C For n>0 a(n) is the maximum element in the continued fraction for P(n)*sqrt(2) where P=A000129 - _Benoit Cloitre_, Jun 19 2005

%H G. C. Greubel, <a href="/A080039/b080039.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: g(t) = (1-t^2+2*t^3)/(1-2*t-2*t^2+2*t^3+t^4).

%F Comments from _Hieronymus Fischer_, 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 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 See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).

%F a(n) = (sqrt(2)+1)^n - (1/2) + (-1)^n*((sqrt(2)-1)^n - (1/2)) for n>0. (End)

%t CoefficientList[Series[(1-t^2+2t^3)/(1-2t-2t^2+2t^3+t^4), {t, 0, 30}], t]

%o (PARI) t='t+O('t^50); Vec((1-t^2+2t^3)/(1-2t-2t^2+2t^3+t^4)) \\ _G. C. Greubel_, Jul 05 2017

%Y Cf. A001622, A006497, A014176, A098316.

%K easy,nonn

%O 0,2

%A Mario Catalani (mario.catalani(AT)unito.it), Jan 21 2003

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Last modified January 16 13:32 EST 2019. Contains 319193 sequences. (Running on oeis4.)