|
| |
|
|
A190949
|
|
Odd Fibonacci numbers with odd index.
|
|
2
|
|
|
|
1, 5, 13, 89, 233, 1597, 4181, 28657, 75025, 514229, 1346269, 9227465, 24157817, 165580141, 433494437, 2971215073, 7778742049, 53316291173, 139583862445, 956722026041, 2504730781961, 17167680177565, 44945570212853, 308061521170129, 806515533049393
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
All prime Fibonacci numbers (A005478) except 2 and 3 are in this sequence. All terms equal 1 (mod 4). The indices of these Fibonacci numbers are 6k-1 or 6k+1.
This sequence can be thought of as two interlocking sequences, each of the form a(n) = 18a(n - 1) - a(n - 2).
Proof that all terms equal 1 (mod 4): From the Lucas 1876 identity Fib(2n+1) = Fib(n)^2 + Fib(n+1)^2 (see Weisstein, formula 60, or page 79 of Koshy), we see that odd-indexed Fibonacci numbers are the sum of two squares. Because a square is 0 or 1 (mod 4), the sum of two squares is 0, 1, or 2 (mod 4). All these terms are odd numbers. Hence, the only possibility is that they are 1 (mod 4). This can also be proved from the recursion formula.
|
|
|
REFERENCES
|
Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001.
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = 18*a(n-2) - a(n-4), with a(1)=1, a(2)=5, a(3)=13, and a(4)=89.
G.f.: x*(1-x)*(1+6*x+x^2)/((1+4*x-x^2)*(1-4*x-x^2)). - Colin Barker, Jun 19 2012
a(n) = (-(2-sqrt(5))^n + (-2-sqrt(5))^n*(-2+sqrt(5)) + 2*(-2+sqrt(5))^n + sqrt(5)*(-2+sqrt(5))^n + (2+sqrt(5))^n)/(2*sqrt(5)) for n>0. - Colin Barker, Mar 27 2016
|
|
|
MATHEMATICA
|
LinearRecurrence[{0, 18, 0, -1}, {1, 5, 13, 89}, 50]
|
|
|
PROG
|
(PARI) Vec(x*(1-x)*(1+6*x+x^2)/((1+4*x-x^2)*(1-4*x-x^2)) + O(x^30)) \\ Colin Barker, Mar 27 2016
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
