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#1 Re: Help Me ! » Same solution for multiple polynomials » 2022-11-20 08:21:52

It's not really important though, never mind

#2 Help Me ! » Same solution for multiple polynomials » 2022-11-20 04:33:02

Dries
Replies: 1

Hey,

I'm just asking this question because I was playing around a little.
But anyhow,

Consider all possible polynomial equations for which the only allowed coefficients are 1 (and 0, but not -1), except for the constant in the polynomial equation.
Is it possible to find a solution, which is not an integer, and which is the solution to more than one of these possible equations?

Thanks

#3 This is Cool » Fractional base for numbers » 2022-11-18 19:26:49

Dries
Replies: 1

So what happens if I use base 3/2 instead of base 10?

Now, this doesn't make any sense. But it is fun to do.
Any number in base 10, say 62, can be written in a form like this: 6 * 10^1 + 2 * 10^0
So I can start building numbers in base 3/2
It starts with 1, as 1 * (3/2)^0
There's no 2, since it would be larger than 10 in base 3/2, since 10 would be 1 * (3/2)^1
However, there's no 11, cause that would be 1 * (3/2)^1 + 1 * (3/2)^0 Which is 2.5 And that's larger than 100 in base 3/2, because 1*(3/2)^2 = 2.25
And so the number line seems to go as follows:
1, 10, 100, 101, 1000, 1001, 1010, 10000, 10001, 10010, 10100, 100000
And the corresponding decimal values are:
1, 1.5, 2.25, 3.25, 3.375, 4.375, 4.875, 5.0625, 6.0625, 6.5625, 7.3125, 7.59375

So we get "whole numbers" that aren't really whole numbers.
And those whole numbers are kind of peculiar. We might add two of these whole numbers, but the result might not be a whole number.
Which kind of makes sense, since 1 ( base 3/2) fits 1.5 times in 10 (base 3/2) and 10 fits 1.5 times in 100.

Now if there were some aliens somewhere in the universe that would only know base 3/2, they might start their number theory with these "whole numbers". But they would find no primes. So it seems. Since the only number that seems to divide 1, is 1 itself.

They would find composite numbers, however. Since, in base 3/2, 10 * 10 = 100 too.
And 10 *101 = 1010 Since 1.5 * 3.25 = 4.875
And 10 * 1010 = 10100 = 100 * 101

But not all numbers are composite.
The sequence of non-composite numbers goes as follows:
1, 101, 1001, 10001, ...

Interesting... An alien mathematician might want to prove a conjecture...

#4 Re: Computer Math » A very inefficient way of approximating the golden ratio » 2022-11-15 02:31:44

Thanks,

I don't know what it is.
It happens too when y approaches 4/3. Sequences get very long.
So for y= 1.3333333 and x= 1 to 999, it found a sequence of length 23,671 at x = 981.
There are 9,799 more odd numbers than even numbers in this sequence.
Some values of y do better than others.
Thanks for the reply though, I was just experimenting, hoping to discover something.
Not easy for me to figure it out. But I had some fun.

#5 Re: Computer Math » A very inefficient way of approximating the golden ratio » 2022-11-14 07:29:35

So in the sequence with x = 819 and y = 1.61803399, I get a sequence length of 26,560 with 15,683 odd numbers and 10,877 even numbers. That's peculiar. :-)

#6 Re: Computer Math » A very inefficient way of approximating the golden ratio » 2022-11-14 02:57:46

It is my post. I lost my password and now I found it.
I don't really understand how these sequences can get so long. Some preference towards odd numbers in the sequences.

#7 Re: Computer Math » Fun math problem » 2021-01-05 07:54:57

Cool. Glad you think so.
I had similar results. The most I've found was 75 steps for 830612 towards 13682.
This time, I removed 0 when it came in front.
This was simpler to implement. I see that you have stopped when you'd loose a digit. Seems cool too.
I have the same result for 103682. It takes 74 steps.

Not sure you would post it to OEIS? All fine by me.

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