 mersenneforum.org My algorithm mimics 2^P-1 with the golden ratio
 Register FAQ Search Today's Posts Mark Forums Read  2018-04-17, 06:59 #1 ONeil   Dec 2017 3×41 Posts My algorithm mimics 2^P-1 with the golden ratio I was fulling 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(2^2-1))^3)+2^2-1   2018-04-17, 07:09   #2
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

23·727 Posts Quote:
 Originally Posted by ONeil I was fulling 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(2^2-1))^3)+2^2-1
I tried with changing the red number to 7 and got 144.768623663...

What did I do wrong?   2018-04-17, 07:12   #3
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

17×83 Posts Quote:
 Originally Posted by retina I tried with changing the red number to 7 and got 144.768623663... What did I do wrong?
In Wolfram's syntax:
Code:
5.2/4^((Sqrt+1)/2*(27+(Sqrt[2^2-1])^3))+2^2-1
it is pretty close to 3, but isn't exactly 3, how you found this expression?

Trivial solution: the exponent of 4 is large: ((sqrt(5)+1)/2*(27+(sqrt(2^2-1))^3))=52.09, so the 5.2/4^exponent is very small, the expression's value will be close to 2^2-1=3.

Last fiddled with by R. Gerbicz on 2018-04-17 at 07:18   2018-04-17, 07:16   #4
ONeil

Dec 2017

3×41 Posts Quote:
 Originally Posted by retina I tried with changing the red number to 7 and got 144.768623663... What did I do wrong?
did you use it at casio?

it works perfectly for me
Attached Thumbnails

Last fiddled with by ONeil on 2018-04-17 at 07:20 Reason: error   2018-04-17, 07:20   #5
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

23·727 Posts Quote:
 Originally Posted by R. Gerbicz In Wolfram's syntax: Code: 5.2/4^((Sqrt+1)/2*(27+(Sqrt[2^2-1])^3))+2^2-1 it is pretty close to 3, but isn't exactly 3
Taking off the final 2^2-1 the expression is zero (or very near to it).

5.2/4^((Sqrt+1)/2*(27+(Sqrt[2^2-1])^3)) ~= 0

Last fiddled with by retina on 2018-04-17 at 07:21   2018-04-17, 07:22   #6
ONeil

Dec 2017

3·41 Posts Quote:
 Originally Posted by R. Gerbicz In Wolfram's syntax: Code: 5.2/4^((Sqrt+1)/2*(27+(Sqrt[2^2-1])^3))+2^2-1 it is pretty close to 3, but isn't exactly 3, how you found this expression? Trivial solution: the exponent of 4 is large: ((sqrt(5)+1)/2*(27+(sqrt(2^2-1))^3))=52.09, so the 5.2/4^exponent is very small, the expression's value will be close to 2^2-1=3.
You have to use the exact algo I sent you to produce exact results.   2018-04-17, 07:24   #7
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

16B816 Posts Quote:
 Originally Posted by ONeil did you use it at casio? it works perfectly for me
It needed an extra set of brackets:

5.2/3.999999999999999^(1.618033988749(27+(sqrt(2^2-1))^3))+2^2-1 ~= 3

So, quite useless IMO.   2018-04-17, 07:26   #8
ONeil

Dec 2017

11110112 Posts Quote:
 Originally Posted by retina It needed an extra set of brackets: 5.2/3.999999999999999^(1.618033988749(27+(sqrt(2^2-1))^3))+2^2-1 ~= 3 So, quite useless IMO.
I just find it to be fascinating that the golden ratio computes this along with the algo.   2018-04-17, 07:29   #9
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

23×727 Posts Quote:
 Originally Posted by ONeil I just find it to be fascinating that the golden ratio computes this along with the algo.
All you have done is this:

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 2018-04-17 at 07:30   2018-04-17, 07:37   #10
ONeil

Dec 2017

3·41 Posts Quote:
 Originally Posted by retina All you have done is this: 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.

Still its interesting because you can edit to get other outputs. Retina what is the difference between an algorithm and a formula?   2018-04-17, 07:40   #11
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

16B816 Posts Quote:
 Originally Posted by ONeil Still its interesting because you can edit to get other outputs.
Not really, it is just 2^n-1, not interesting at all unless you consider the trailing 10^-31 (which your calculator hid from you). You are basically saying that 2^n-1 equals 2^n-1.
Quote:
 Originally Posted by ONeil Retina what is the difference between an algorithm and a formula?    Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post xilman Reading 2 2015-11-07 02:18 davieddy Puzzles 9 2010-07-04 22:21 davar55 Puzzles 2 2008-02-13 03:20 mfgoode Puzzles 1 2007-01-31 16:26 mfgoode Math 28 2004-05-31 10:40

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