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A080039 a(n) = floor((1+sqrt(2))^n). 10
1, 2, 5, 14, 33, 82, 197, 478, 1153, 2786, 6725, 16238, 39201, 94642, 228485, 551614, 1331713, 3215042, 7761797, 18738638, 45239073, 109216786, 263672645, 636562078, 1536796801, 3710155682, 8957108165, 21624372014, 52205852193 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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

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

LINKS

G. C. Greubel, Table of n, a(n) for n = 0..1000

FORMULA

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

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).

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.

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

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

MATHEMATICA

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

PROG

(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

CROSSREFS

Cf. A001622, A006497, A014176, A098316.

Sequence in context: A096772 A090803 A018015 * A265226 A299164 A131408

Adjacent sequences:  A080036 A080037 A080038 * A080040 A080041 A080042

KEYWORD

easy,nonn

AUTHOR

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

STATUS

approved

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Last modified December 13 10:42 EST 2018. Contains 318086 sequences. (Running on oeis4.)