20180417, 06:59  #1 
Dec 2017
2^{4}·3·5 Posts 
My algorithm mimics 2^P1 with the golden ratio
I was messing around with the golden ratio and some other numbers at
https://keisan.casio.com/calculator and I produced this algorithm. The input is in red. It will find mersenne numbers 5.2/3.999999999999999^1.618033988749(27+(sqrt(2^21))^3)+2^21 Last fiddled with by ONeil on 20201104 at 20:39 
20180417, 07:09  #2  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6291_{10} Posts 
Quote:
What did I do wrong? 

20180417, 07:12  #3  
"Robert Gerbicz"
Oct 2005
Hungary
2727_{8} Posts 
Quote:
Code:
5.2/4^((Sqrt[5]+1)/2*(27+(Sqrt[2^21])^3))+2^21 Trivial solution: the exponent of 4 is large: ((sqrt(5)+1)/2*(27+(sqrt(2^21))^3))=52.09, so the 5.2/4^exponent is very small, the expression's value will be close to 2^21=3. Last fiddled with by R. Gerbicz on 20180417 at 07:18 

20180417, 07:16  #4  
Dec 2017
11110000_{2} Posts 
Quote:
it works perfectly for me Last fiddled with by ONeil on 20180417 at 07:20 Reason: error 

20180417, 07:20  #5  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
14223_{8} Posts 
Quote:
5.2/4^((Sqrt[5]+1)/2*(27+(Sqrt[2^21])^3)) ~= 0 Last fiddled with by retina on 20180417 at 07:21 

20180417, 07:22  #6  
Dec 2017
2^{4}·3·5 Posts 
Quote:


20180417, 07:24  #7 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1100010010011_{2} Posts 

20180417, 07:26  #8 
Dec 2017
2^{4}×3×5 Posts 

20180417, 07:29  #9  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{3}·233 Posts 
Quote:
c + 2^n  1, where c is ~= 10^31 So it might as well be: 0 + 2^n  1, which is just 2^n  1 ETA: It isn't an "algo", it is a formula. Last fiddled with by retina on 20180417 at 07:30 

20180417, 07:37  #10  
Dec 2017
2^{4}·3·5 Posts 
Quote:
Still its interesting because you can edit to get other outputs. Retina what is the difference between an algorithm and a formula? 

20180417, 07:40  #11 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
3^{3}×233 Posts 
Not really, it is just 2^n1, not interesting at all unless you consider the trailing 10^31 (which your calculator hid from you). You are basically saying that 2^n1 equals 2^n1.

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