|
| |
|
|
A077925
|
|
Expansion of 1/((1-x)*(1+2*x)).
|
|
43
|
|
|
|
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)
Index entries related to Gaussian binomial coefficients.
Index entries for Lucas sequences
|
|
|
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
|
|
|
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,changed
|
|
|
AUTHOR
|
N. J. A. Sloane, Nov 17 2002
|
|
|
STATUS
|
approved
|
| |
|
|
