OFFSET
0,2
COMMENTS
The limit as n -> infinity of a(n+1)/a(n) is (1+sqrt(5))*(2+sqrt(5))/2 = 6.8541019662...
Old name was: Difference between two Fibonacci cycles A000045 (three's cycle minus two's cycle).
LINKS
Index entries for linear recurrences with constant coefficients, signature (11,-28,-3,1)
FORMULA
a(n) = F(4n+4)/F(4)-F(3n+3)/F(3) = (2*F(4n+4)-3*F(3n+3))/6, where F=A000045.
G.f.: x*(2*x-3) / ((x^2-7*x+1)*(x^2+4*x-1)). - Colin Barker, Mar 15 2013
MATHEMATICA
a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] b = Table[a[4*(n + 1)]/a[4], {n, 0, 25}]; c = Table[a[3*(n + 1)]/a[3], {n, 0, 25}]; aout = b - c
LinearRecurrence[{11, -28, -3, 1}, {0, 3, 31, 257}, 20] (* Harvey P. Dale, Jun 09 2022 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Roger L. Bagula, Feb 20 2006
EXTENSIONS
Edited by Bruno Berselli, Mar 15 2013
Put formula as clearer name, Joerg Arndt, Mar 15 2013
STATUS
approved