|
| |
|
|
A102875
|
|
Let f(n) = n+2 if n == 1 mod 3, = n if n == 2 mod 3, = n-2 if n == 0 mod 3; then a(n) = Fibonacci(f(n)).
|
|
0
|
|
|
|
0, 2, 1, 1, 8, 5, 3, 34, 21, 13, 144, 89, 55, 610, 377, 233, 2584, 1597, 987, 10946, 6765, 4181, 46368, 28657, 17711, 196418, 121393, 75025, 832040, 514229, 317811, 3524578, 2178309, 1346269, 14930352, 9227465, 5702887, 63245986, 39088169
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
In other words, split the Fibonacci numbers into groups of three and reverse each group.
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f.: x*(-2-x-x^2-x^4+x^5) / ( (x^2+x-1)*(x^4-x^3+2*x^2+x+1) ). - R. J. Mathar, Mar 08 2011
|
|
|
MATHEMATICA
|
fib[n_Integer?Positive] := fib[n] = fib[n - 1] + fib[n - 2] fib[0] = 0; fib[1] = 1 f[n_] = If[Mod[n, 3] == 1, n + 2, If[Mod[n, 3] == 0, n - 2, n]] a = Table[fib[f[n]], {n, 1, 200}]
Flatten[Join[{0}, Reverse/@Partition[Fibonacci[Range[42]], 3]]] (* or *) Join[{0}, LinearRecurrence[{0, 0, 4, 0, 0, 1}, {2, 1, 1, 8, 5, 3}, 40]] (* Harvey P. Dale, Mar 26 2015 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
