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Old 2018-04-17, 06:59   #1
ONeil
 
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Arrow 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
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Old 2018-04-17, 07:09   #2
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Quote:
Originally Posted by ONeil View Post
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?
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Old 2018-04-17, 07:12   #3
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Quote:
Originally Posted by retina View Post
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[5]+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
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Old 2018-04-17, 07:16   #4
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Quote:
Originally Posted by retina View Post
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
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Last fiddled with by ONeil on 2018-04-17 at 07:20 Reason: error
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Old 2018-04-17, 07:20   #5
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Quote:
Originally Posted by R. Gerbicz View Post
In Wolfram's syntax:
Code:
5.2/4^((Sqrt[5]+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[5]+1)/2*(27+(Sqrt[2^2-1])^3)) ~= 0

Last fiddled with by retina on 2018-04-17 at 07:21
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Old 2018-04-17, 07:22   #6
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Quote:
Originally Posted by R. Gerbicz View Post
In Wolfram's syntax:
Code:
5.2/4^((Sqrt[5]+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.
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Old 2018-04-17, 07:24   #7
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Quote:
Originally Posted by ONeil View Post
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.
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Old 2018-04-17, 07:26   #8
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Quote:
Originally Posted by retina View Post
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.
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Old 2018-04-17, 07:29   #9
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Quote:
Originally Posted by ONeil View Post
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
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Old 2018-04-17, 07:37   #10
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Quote:
Originally Posted by retina View Post
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?
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Old 2018-04-17, 07:40   #11
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Quote:
Originally Posted by ONeil View Post
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:
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Retina what is the difference between an algorithm and a formula?
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