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 A077925 Expansion of 1/((1-x)*(1+2*x)). 49
 1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, -1365, 2731, -5461, 10923, -21845, 43691, -87381, 174763, -349525, 699051, -1398101, 2796203, -5592405, 11184811, -22369621, 44739243, -89478485, 178956971, -357913941, 715827883, -1431655765, 2863311531, -5726623061 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS a(n+1) is the reflection of a(n) through a(n-1) on the numberline. - Floor van Lamoen, Aug 31 2004 If a zero is added as the (new) a(0) in front, the sequence represents the inverse binomial transform of A001045. Partial sums are in A077898. - R. J. Mathar, Aug 30 2008 a(n) = A077953(2*n+3). - Reinhard Zumkeller, Oct 07 2008 Related to the Fibonacci sequence by an INVERT transform: if A(x) = 1+x^2*g(x) is the generating function of the a(n) prefixed with 1, 0, then 1/A(x) = 2+(x+1)/(x^2-x+1) is the generating function of 1, 0, -1, 1, -2, 3, ..., the signed Fibonacci sequence A000045 prefixed with 1. - Gary W. Adamson, Jan 07 2011 Also: Gaussian binomial coefficients [n+1,1], or q-integers, for q=-2, diagonal k=1 in the triangular (or column r=1 in the square) array A015109. - M. F. Hasler, Nov 04 2012 With a leading zero, 0, 1, -1, 3, -5, 11, -21, 43, -85, 171, -341, 683, ... we obtain the Lucas U(-1,-2) sequence. - R. J. Mathar, Jan 08 2013 Let m = a(n). Then 18*m^2 - 12*m + 1 = A000225(2n+3). - Roderick MacPhee, Jan 17 2013 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Wikipedia, Lucas sequence Index entries for linear recurrences with constant coefficients, signature (-1,2) FORMULA G.f.: 1/(1+x-2*x^2). a(n) = (1-(-2)^(n+1))/3. - Vladeta Jovovic, Apr 17 2003 a(n) = Sum_{k=0..n} (-2)^k. - Paul Barry, May 26 2003 a(n+1) - a(n) = A122803(n). - R. J. Mathar, Aug 30 2008 a(n) = Sum_{k=0..n} A112555(n,k)*(-2)^k. - Philippe Deléham, Sep 11 2009 a(n) = A082247(n+1) - 1. - Philippe Deléham, Oct 07 2009 G.f.: Q(0)/(3*x), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k + 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013 G.f.: Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k-1 + 2*x)/( x*(4*k+1 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013 E.g.f.: (2*exp(-2*x) + exp(x))/3. - Ilya Gutkovskiy, Nov 12 2016 a(n) = A086893(n+2) - A061547(n+3), n >= 0. - Yosu Yurramendi, Jan 16 2017 a(n) = (-1)^n*A001045(n+1). - M. F. Hasler, Feb 13 2020 EXAMPLE G.f. = 1 - x + 3*x^2 - 5*x^3 + 11*x^4 - 21*x^5 + 43*x^6 - 85*x^7 + ... MAPLE a:=n->sum ((-2)^j, j=0..n): seq(a(n), n=0..35); # Zerinvary Lajos, Dec 16 2008 MATHEMATICA CoefficientList[Series[(1 - x)^(-1)/(1 + 2 x), {x, 0, 50}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 20 2011 *) PROG (Sage) [gaussian_binomial(n, 1, -2) for n in range(1, 35)] # Zerinvary Lajos, May 28 2009 (MAGMA) [(1-(-2)^(n+1))/3: n in [0..40]]; // Vincenzo Librandi, Jun 21 2011 (PARI) a(n)=(1+(-2)^n*2)/3 \\ Charles R Greathouse IV, Jun 21 2011 CROSSREFS Cf. A001045 (unsigned version). Cf. A014983, A014985, A014986. - Zerinvary Lajos, Dec 16 2008 Cf. A232600, A232601, A232602. Sequence in context: A328284 A167167 A001045 * A152046 A283642 A284426 Adjacent sequences:  A077922 A077923 A077924 * A077926 A077927 A077928 KEYWORD sign,easy AUTHOR N. J. A. Sloane, Nov 17 2002 STATUS approved

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Last modified August 8 23:57 EDT 2022. Contains 356016 sequences. (Running on oeis4.)