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Martin Hopf
Martin Hopf
6:47 AM
Write a neat algorithm that displays as much as possible digits of the Feigenbaum constant.
Sounds simple but:
{
Feigenbaum(it)=
my(a1 = 1.0, a2 = 0.0, d1 = 3.2, jt = it-3, x, y);
for(i = 2, it,
a = a1 + (a1 - a2) / d1;
for( j = 1, jt,
x = 0.;
y = 0.;
for (k = 1, 2^i,
y = 1 - 2 * y * x;
x = a - sqr(x)
);
a -= x/y
);
d = (a1 - a2)/ (a - a1);
printf("%d\t%f\n",i ,d);
d1 = d;
a2 = a1;
a1 = a;
)
};
? Feigenbaum(25)
2 3.2185114220380879122705045307428132560
3 4.3856775985683390857449485687755223456
4 4.6009492765380753578116946986238349857
5 4.6551304953919801364862549958568988073
6 4.6661119478285713883312136967117765828
7 4.6685485814468409480445436801481451762
8 4.6690606606482682391325998226302752086
9 4.6691715553795113888860046098975782017
10 4.6691951560300171740211088011913071831
11 4.6692002290868564979383537810028739836
12 4.6692013132942041711647549411986485821
13 4.6692015457809067075060581098195881882
14 4.6692015955374939102924706402488974182
15 4.6692016061981521577238310960356646083
? ##
*** last result: cpu time 7min, 10,549 ms, real time 7min, 10,581 ms.

From OEIS
4.669201609102990671853203820466201617258185577475768632745651343004134...

It needs a long time to do 25 iterations which gives about 16 valid digits of the constant.

Can we get much more precision?
 
 
5 hours later…
Peter
Peter
11:55 AM
@MartinHopf I am suprised that one can calculate this constant at all with such a high precision. I heard of this (and another closely related) constant but was not aware of algorithms to calculate them.
 
Peter
Peter
12:26 PM
By the way, you are invited to join to the ec-prime search (or the doublecheck part) or to my aborted $5^n+6^n+10^n$ - prime search or factor a number with $2022$ involved
 

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