|
Search: id:A069921
|
|
|
| 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. |
|
+0
6
|
|
| 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; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
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.yu), May 06 2002
a(n) = F(n)^2 + F(n+2)^2. - Ron Knott (enquiry(AT)ronknott.com), Aug 02 2004
|
|
MATHEMATICA
|
a[n_] := 3Fibonacci[n]Fibonacci[n+2]+(-1)^n
|
|
CROSSREFS
|
Cf. A069959-A069963.
Adjacent sequences: A069918 A069919 A069920 this_sequence A069922 A069923 A069924
Sequence in context: A093029 A105505 A005514 this_sequence A053818 A133629 A048010
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
Benoit Cloitre (benoit7848c(AT)orange.fr), May 05 2002
|
|
EXTENSIONS
|
Edited by Dean Hickerson (dean(AT)math.ucdavis.edu), May 08 2002
|
|
|
Search completed in 0.002 seconds
|
