|
| |
|
|
A069961
|
|
Define C(n) by the recursion C(0)=4*I where I^2=-1, C(n+1)=1/(1+C(n)); then a(n)=4*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.
|
|
2
| |
|
|
1, 17, 20, 73, 169, 464, 1193, 3145, 8212, 21521, 56321, 147472, 386065, 1010753, 2646164, 6927769, 18137113, 47483600, 124313657, 325457401, 852058516, 2230718177, 5840095985, 15289569808, 40028613409, 104796270449
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,2
|
|
|
COMMENTS
| If we define C(n) with C(0)=I then Im(C(n))=1/F(2n+1) where F(k) are the Fibonacci numbers.
C(n) = (F(n)+F(n-1)*C(0))/(F(n+1)+F(n)*C(0)) = (F(n)*(F(n+1)+16*F(n-1))+(-1)^n*4*I)/(F(n+1)^2+16*F(n)^2).
|
|
|
FORMULA
| a(n) = 16 F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number.
|
|
|
MATHEMATICA
| a[n_] := 16Fibonacci[n]^2+Fibonacci[n+1]^2; Array[a, 30, 0]
16First[#]^2+Last[#]^2&/@Partition[Fibonacci[Range[0, 30]], 2, 1] (* From Harvey P. Dale, Nov 08 2011 *)
|
|
|
CROSSREFS
| Cf. A069921, A069959, A069960, A069962, A069963.
Sequence in context: A081643 A146169 A045020 * A140146 A039502 A039505
Adjacent sequences: A069958 A069959 A069960 * A069962 A069963 A069964
|
|
|
KEYWORD
| easy,nonn
|
|
|
AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 28 2002
|
|
|
EXTENSIONS
| Edited by Dean Hickerson (dean.hickerson(AT)yahoo.com), May 08 2002
|
| |
|
|
