OFFSET
0,2
COMMENTS
Ratio of successive terms approaches sqrt(5) + 1.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Harry J. Smith)
Index entries for linear recurrences with constant coefficients, signature (2,4).
FORMULA
For n >= 1, a(n) = 2^(n-1)*Fibonacci(n+3). - Vladeta Jovovic, Oct 25 2003
G.f.: (1 + x)/(1 - 2*x - 4*x^2). - R. J. Mathar, Feb 06 2010
a(n) = (1/2)*(1+sqrt(5))^n + (1/5)*(1+sqrt(5))^n*sqrt(5) - (1/5)*sqrt(5)*(1-sqrt(5))^n + (1/2)*(1-sqrt(5))^n. - Alexander R. Povolotsky, Aug 15 2010
It follows that a(n) is the nearest integer to (and is increasingly close to) (1/2 + 1/sqrt(5))*(1+sqrt(5))^n. - N. J. A. Sloane, Aug 10 2012
a(n) = M^n(1, 1), with the matrix M= [[3, 1], [1, -1]]. Proof by Cayley-Hamilton, using S(n, -I) = (-I)^n*F(n+1), and S = A049310 and F = A000045. Motivated by A319053. - Wolfdieter Lang, Oct 08 2018
EXAMPLE
As the INVERT transform of A006138, (1, 2, 5, 11, 26, 59, ...); a(4) = 104 = (26, 11, 5, 2, 1) dot (1, 1, 3, 10, 32) = (26 + 11 + 15 + 20 + 32).
MAPLE
a := proc(n) option remember: if n=0 then RETURN(1) fi: if n=1 then RETURN(2) fi: 2*a(n-1) + 4*a(n-2); end: for n from 1 to 50 do printf(`%d, `, a(n)+a(n-1)) od:
f:=n-> simplify(expand((1/2)*(1+sqrt(5))^n + (1/5)*(1+sqrt(5))^n*sqrt(5) - (1/5)*sqrt(5)*(1-sqrt(5))^n + (1/2)*(1 -sqrt(5))^n )); # N. J. A. Sloane, Aug 10 2012
MATHEMATICA
a[n_]:=(MatrixPower[{{1, 5}, {1, 1}}, n].{{2}, {1}})[[2, 1]]; Table[a[n], {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{2, 4}, {1, 3}, 100] (* G. C. Greubel, Feb 18 2017 *)
PROG
(PARI) { for (n=0, 200, if (n>1, a=2*a1 + 4*a2; a2=a1; a1=a, if (n, a=a1=2, a=a2=1)); if (n, write("b063782.txt", n, " ", a + a2)) ) } \\ Harry J. Smith, Aug 31 2009
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Klaus E. Kastberg (kastberg(AT)hotkey.net.au), Aug 17 2001
EXTENSIONS
More terms from James A. Sellers, Sep 25 2001
Edited (new offset, new initial term, etc.) by N. J. A. Sloane, Aug 19 2010
STATUS
approved