|
| |
|
|
A138573
|
|
a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=2, a(3)=5.
|
|
5
|
|
|
|
0, 1, 2, 5, 16, 45, 130, 377, 1088, 3145, 9090, 26269, 75920, 219413, 634114, 1832625, 5296384, 15306833, 44237570, 127848949, 369490320, 1067846845, 3086134658, 8919094697, 25776662080, 74495936025, 215297250946, 622220603405
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
This is a divisibility sequence; that is, if n divides m, then a(n) divides a(m). - T. D. Noe, Dec 23 2008
Case P1 = 2, P2 = -4, Q = 1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 04 2014
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = round(w^n/2/sqrt(5)) where w = (1+r)/(1-r) = 2.89005363826396... and r = sqrt(sqrt(5)-2) = 0.485868271756...; for n >= 3, a(n) = A071101(n+3).
G.f.: -x*(x-1)*(1+x)/(1 - 2*x - 2*x^2 - 2*x^3 + x^4). - R. J. Mathar, Jun 03 2009
Define a Lucas sequence {U(n)} in the ring of Gaussian integers by the recurrence U(n) = (1 + i)*U(n-1) + U(n-2) with U(0) = 0 and U(1) = 1. Then a(n) = |U(n)|^2.
Let a, b denote the zeros of x^2 - (1 + i)*x - 1 and c, d denote the zeros of x^2 - (1 - i)*x - 1.
Then a(n) = (a^n - b^n)*(c^n - d^n)/((a - b)*(c - d)).
a(n) = (alpha(1)^n + beta(1)^n - alpha(2)^n - beta(2)^n)/(2*sqrt(5)), where alpha(1), beta(1) are the roots of x^2 - ( 1 + sqrt(5))*x + 1 = 0, and alpha(2), beta(2) are the roots of x^2 - (1 - sqrt(5))*x + 1 = 0.
The o.g.f. is the Hadamard product of the rational functions x/(1 - (1 + i)x - x^2) and x/(1 - (1 - i)x - x^2). (End)
a(n) = (1/sqrt(5))*(T(n,phi) - T(n,-1/phi)), where phi = 1/2*(1 + sqrt(5)) is the golden ratio and T(n,x) denotes the Chebyshev polynomial of the first kind. Compare with the Fibonacci numbers, A000045, whose terms are given by the Binet formula 1/sqrt(5)*( phi^n - (-1/phi)^n ).
a(n) = top right (or bottom left) entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, 1; 1, 1]; the off-diagonal elements of M^n give the sequence of Fibonacci numbers. Bottom right entry of the matrix T(n, M) gives A138574. See the remarks in A100047 for the general connection between Chebyshev polynomials and linear divisibility sequences of the fourth order. (End)
a(n) = (((phi + sqrt(phi))^n + (phi - sqrt(phi))^n)/2 - (-1)^n * cos(n*arctan(sqrt(phi))))/sqrt(5), where phi=(1+sqrt(5))/2. - Vladimir Reshetnikov, May 11 2016
|
|
|
MAPLE
|
seq(coeff(series((x*(1-x)*(x+1))/(1-2*x-2*x^2-2*x^3+x^4), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Sep 12 2018
|
|
|
MATHEMATICA
|
Round@Table[(((GoldenRatio + Sqrt[GoldenRatio])^n + (GoldenRatio - Sqrt[GoldenRatio])^n)/2 - (-1)^n Cos[n ArcTan[Sqrt[GoldenRatio]]])/Sqrt[5], {n, 0, 20}] (* or *) LinearRecurrence[{2, 2, 2, -1}, {0, 1, 2, 5}, 20] (* Vladimir Reshetnikov, May 11 2016 *)
CoefficientList[Series[-x*(x-1)*(1+x)/(1-2*x-2*x^2-2*x^3+x^4), {x, 0, 20}], x] (* Stefano Spezia, Sep 12 2018 *)
|
|
|
PROG
|
(PARI) x='x+O('x^50); concat([0], Vec(x*(1-x)*(1+x)/(1 -2*x -2*x^2 -2*x^3 +x^4))) \\ G. C. Greubel, Aug 08 2017
(GAP) a:=[0, 1, 2, 5];; for n in [5..30] do a[n]:=2*a[n-1]+2*a[n-2]+2*a[n-3]-a[n-4]; od; a; # Muniru A Asiru, Sep 12 2018
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
