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A123335 a(n) = -2*a(n-1)+a(n-2) for n>1, a(0)=1, a(1)=-1. 10
1, -1, 3, -7, 17, -41, 99, -239, 577, -1393, 3363, -8119, 19601, -47321, 114243, -275807, 665857, -1607521, 3880899, -9369319, 22619537, -54608393, 131836323, -318281039, 768398401, -1855077841, 4478554083, -10812186007, 26102926097, -63018038201, 152139002499 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Inverse binomial transform of A077957 .

The inverse of the g.f. is 3-x-2/(1+x) which generates 1, 1, -2, +2, -2, +2,... (-2, +2 periodically continued). - Gary W. Adamson, Jan 10 2011

Pisano period lengths:  1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12,... - R. J. Mathar, Aug 10 2012

a(n) is the rational part of the Q(sqrt(2)) integer (sqrt(2) - 1)^n = a(n) + A077985(n-1)*sqrt(2), with A077985(-1) = 0. - Wolfdieter Lang, Dec 07 2014

3^n*a(n) = A251732(n) gives the rational part of the integer in Q(sqrt(2)) giving the length of a variant of Lévy's C-curve at iteration step n. - Wolfdieter Lang, Dec 07 2014

LINKS

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (-2,1).

FORMULA

a(n) = (-1)^n*A001333(n).

G.f.: (1+x)/(1+2*x-x^2).

a(n) = 1/2*((-1-sqrt(2))^n+(-1+sqrt(2))^n). [Paolo P. Lava, Nov 19 2008]

a(n) = A077985(n)+A077985(n-1). - R. J. Mathar, Mar 28 2011

G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 19 2013

MATHEMATICA

LinearRecurrence[{-2, 1}, {1, -1}, 40] (* Harvey P. Dale, Nov 03 2011 *)

CROSSREFS

Cf. A001333, A077985, A251732.

Sequence in context: A077851 A089737 A001333 * A078057 A089742 A187258

Adjacent sequences:  A123332 A123333 A123334 * A123336 A123337 A123338

KEYWORD

sign,easy

AUTHOR

Philippe Deléham, Jun 27 2007

EXTENSIONS

Corrected by N. J. A. Sloane, Oct 05 2008

STATUS

approved

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Last modified December 10 07:27 EST 2016. Contains 278993 sequences.