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Template:Sequence of the Day for September 27

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Intended for: September 27, 2014

Timetable

  • First draft entered by Alonso del Arte on September 1, 2013
  • Draft reviewed by Daniel Forgues on October 02, 2016
  • Draft to be approved by August 27, 2014
Yesterday's SOTD * Tomorrow's SOTD

The line below marks the end of the <noinclude> ... </noinclude> section.



A192038: The Fibonacci “logarithm” for 4.*

4.5491125565...
What is the ten millionth Fibonacci number? Without the Binet formula, answering that question would require computing ten million Fibonacci numbers. Thanks to the Binet formula,
F (n) =
ϕn  −  ( − 1)n ϕ  −  n
2  5
, where
ϕ =
1 +
2  5
2
is the golden ratio, we can just plug in
n = 10 7
and get the
n
th Fibonacci number with just a few operations instead of ten million or so additions. The thought then occurs: Must we plug in an integer? What happens if we put in any real number? Defining
f  (x) :=
ϕ  x  −  cos (π  x) ϕ  −  x
2  5
(a slight variation of the Binet formula, Clark Kimberling tells us)** does not readily indicate any impediments to plugging a real number, and so, for example, we have
f  (2 
2  5
 )   ≈   3.84257859447.
The next question then is: Can we find a value of
x
that will give us an integer that is not actually a Fibonacci number, like, say, 4? Indeed we can. Setting
x   ≈   4.5491125565
we obtain
f  (x) = 4
. That’s one solution, there are others, which are all negative numbers.

_______________

* Since the Fibonacci function
f  (x), x ∈ ℝ ,
is strictly increasing only from
x = 1.67668837258...
(A171909), the Fibonacci “logarithm” has a unique value for
x   ≥   1.67668837258...
.

** Merve Özvatan and Oktay K. Pashaev, Generalized Fibonacci Sequences and Binet–Fibonacci Curves (July 2017), arXiv:1707.09151. (Cf. §5.1 Fibonacci Numbers of Real argument and §5.3 Fibonacci Spirals.)

Notes