|
| |
|
|
A069921
|
|
Define C(n) by the recursion C(0)=1+I where I^2=-1, C(n+1)=1/(1+C(n)); then a(n)=(-1)^n/Im(C(n)) where Im(z) is the imaginary part of the complex number z.
|
|
5
| |
|
|
1, 5, 10, 29, 73, 194, 505, 1325, 3466, 9077, 23761, 62210, 162865, 426389, 1116298, 2922509, 7651225, 20031170, 52442281, 137295677, 359444746, 941038565, 2463670945, 6449974274, 16886251873, 44208781349, 115740092170
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,2
|
|
|
COMMENTS
| C(n) = (F(n)+F(n-1)*C(0))/(F(n+1)+F(n)*C(0)) = (3*F(n)*F(n+1)+(-1)^n*(1+I))/(3*F(n)*F(n+2)+(-1)^n)
|
|
|
FORMULA
| a(n) = 3 F(n) F(n+2) + (-1)^n, where F(n) = A000045(n) is the n-th Fibonacci number. Also, a(n)=ceiling(3/5*(g/2)^(n+1))-(1+(-1)^n)/2 with g=3+sqrt(5).
a(n) = 2*a(n-1)+2*a(n-2)-a(n-3). G.f.: (1+3*x-2*x^2)/(1-2*x-2*x^2+x^3). - Vladeta Jovovic (vladeta(AT)eunet.rs), May 06 2002
a(n) = F(n)^2 + F(n+2)^2. - Ron Knott (enquiry(AT)ronknott.com), Aug 02 2004
Essentially the same sequence is produced by the rule a(n) = (A000045(n)^2+A000032(n)^2)/2. [From Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Nov 17 2008]
|
|
|
MATHEMATICA
| a[n_] := 3Fibonacci[n]Fibonacci[n+2]+(-1)^n
(*A000045*) F[n_] := (((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n)/Sqrt[5]; (*A000032*) L[n_] := ((1 + Sqrt[5])/2)^n + ((1 - Sqrt[5])/2)^n; Table[FullSimplify[ExpandAll[(F[n]^2 + L[n]^2)/2]], {n, 0, 50}] [From Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Nov 17 2008]
LinearRecurrence[{2, 2, -1}, {1, 5, 10}, 70] (* From Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)
|
|
|
CROSSREFS
| Cf. A069959-A069963.
Cf. A059929, A000032, A000045 [From Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Nov 17 2008]
Sequence in context: A093029 A105505 A005514 * A053818 A133629 A156302
Adjacent sequences: A069918 A069919 A069920 * A069922 A069923 A069924
|
|
|
KEYWORD
| easy,nonn,changed
|
|
|
AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), May 05 2002
|
|
|
EXTENSIONS
| Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), May 08 2002
|
| |
|
|
