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Fibonacci[1,000,000,000] contains 208,987,640 decimal digits (was: Fibonachi[5,000,000] contains 1044938 decimal digits)
*To*: mathgroup at smc.vnet.net
*Subject*: [mg49649] Fibonacci[1,000,000,000] contains 208,987,640 decimal digits (was: Fibonachi[5,000,000] contains 1044938 decimal digits)
*From*: "Michael Taktikos" <michael.taktikos at hanse.net>
*Date*: Mon, 26 Jul 2004 04:01:53 -0400 (EDT)
*References*: <7f4ffu$6dj$1@nnrp1.dejanews.com>, <Mo2OZGACWlF3Ewez@raos.demon.co.uk>, <7g0qsd$dr9@smc.vnet.net> <cdvm0e$7hv$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Alex Vinokur <alexvn at bigfoot.com> wrote:
> >> >Several large Fibonacci numbers were calculated using only
> >> >the well-known explicit formula:
> >> > Fib (0) = 0, Fib (1) = 1,
> >> > Fib (n) = Fib (n-1) + Fib (n-2), n >= 2
> >> >All the (decimal) digits of these numbers were obtained.
[...]
> >> >But to get ALL digits of the large Fibonacci number is very
> advisable one.
> [C++ code, snip]
Thanks for your code.
Until now, the fastest way to get with "Mathematica-alone" all digits of
large Fibonacci numbers seems to be that of Roman Maeder:
********************************************************
fibmaeder[n_] :=
Module[{r11 = 1, r12 = 0, r22 = 1, digits = IntegerDigits[n-1, 2], i, t},
Do[ If[ digits[[i]] == 1,
{r11, r22} = {r11(r11 + 2r12), r12(r11 + r22)};
r12 = r11 - r22
, t = r12(r11 + r22);
{r11, r12} = {r11(r11 + 2r12) - t, t};
r22 = r11 - r12
],
{i, Length[digits]-1}
];
If[ digits[[-1]] == 1,
r11(r11 + 2r12),
r11(r11 + r22) - (-1)^((n-1)/2)
]
]
********************************************************
If you don't need all the digits, but only the number of the contained
decimal digits,
you can use the function of Binet with an appropriate precision:
In[1]:=fiboBinet[n_]:=Module[{gr=GoldenRatio^n}, N[(gr+1/gr)/Sqrt[5],1000]]
In[2]:=Timing[fiboBinet[10^9]]
(gives with my old AMD 700 MHz the answer {0.6 Second,
7.9...*10^208987639} )
Now I detected an other function, which seems to be faster than Binet's one:
fiboNew(n) = g^n/(2g-1) with g = GoldenRatio.
Let's try it:
In[3]:=fiboNew[n_]:=N[(GoldenRatio^n)/(2*GoldenRatio-1),1000];
In[4]:=Timing[fiboNew[10^9]]
It gives the answer {0.1 Second, 7.9...*10^208987639}
By searching with Google and looking in your "Bronstein" you will not find
it published,
therefore I ask the group: is this the first publication or did I only
discovered the wheel again?
I believe to have a proof that this function gives allways the same answer
as Binet's one,
now I'm looking if it contains bugs.
Greetings from Hamburg,
Michael Taktikos
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