Re: Fibonacci[1,000,000,000] contains 208,987,640 decimal digits

From: Michael Taktikos (michael.taktikos_at_hanse.net)
Date: 08/10/04 

Date: Tue, 10 Aug 2004 21:23:15 +0200

"Klueless" <klueless@worldnet.att.net> wrote:

> "Michael Taktikos" <michael.taktikos@hanse.net> wrote in message
news:2npfirF385mqU1@uni-berlin.de...
> > "Ed Pegg Jr" <edpegg@gmail.com> wrote in comp.soft-sys.math.mathematica:
> > > For finding the number of digits in any sufficently large Fibonacci
> > > number in base 10, let k = (ArcCsch[2])/Log[10]. The number of
> > > digits in Fibonacci[n] is Round[n k].

To summarize a bit: The result, that Fibonacci[10^9] has 208,987,640 digits
(and hence the value in Sloane's A068070 is wrong), was found by my in
Mathematica implemented Fibonacci function

fiboappr[n_]:=N[(GoldenRatio^n)/(2*GoldenRatio-1),1000]

and published July, 26, 2004 in the group comp.soft-sys.math.mathematica.

On August, 9, 2004, Mr. Pegg wrote the above mail confirming that the same
result was found by his function
Round[n*k] with k = (ArcCsch[2])/Log[10], which he believed would give the
number of digits
of Fibonacci[n]. Unfortunatelly, this confirmity was only accidentally,
because his function gives sometimes
wrong results.

In the meantime, Mr. Pegg posted a correct sequence as Sloane's A097348, in
which he uses the formula
FibonacciDigits[n_] := Ceiling[(2*n*ArcCsch[2] - Log[5])/Log[100]]

K> This formula:
K>
K> Round[(ArcCsch[2]*n+Log[2]/2)/Log[10]] (n>=2)
K>
K> should be more reliable, but may be still not perfect.

Looks good, in the first million n (from n=2 on) it gives the same results
as the FibonacciDigits above
and after transforming to

Round[(2*n*ArcCsch[2] + Log[2])/Log[100]] (n >=2)

it works faster than FibonacciDigits.

Greetings from Hamburg,

Michael Taktikos