For any odd number x, there exists another odd number y, such that the
product of x and y expressed in binary is a contiguous sequence of 1s,
though I'll express them in hex.

For example
3 * 1 = 3
5 * 3 = f
7 * 9 = 3f
9 * 7 = 3f
11 * 93 = 3ff
13 * 315 = fff

97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF

Sylvia.

Sylvia Else wrote:

> For any odd number x, there exists another odd number y, such that the
> product of x and y expressed in binary is a contiguous sequence of 1s,
> though I'll express them in hex. For example
> 3 * 1 = 3 5 * 3 = f 7 * 9 = 3f 9 * 7 = 3f
11 * 93
> = 3ff 13 * 315 = fff
>
> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF

nonsense, 11 * 3 = 33 = 21h

On 07-Nov-21 9:30 am, Gregor Bicha wrote:
> Sylvia Else wrote:
>
>> For any odd number x, there exists another odd number y, such that the
>> product of x and y expressed in binary is a contiguous sequence of 1s,
>> though I'll express them in hex. For example
>> 3 * 1 = 3 5 * 3 = f 7 * 9 = 3f 9 * 7 = 3f
> 11 * 93
>> = 3ff 13 * 315 = fff
>>
>> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>
> nonsense, 11 * 3 = 33 = 21h
>

I didn't say that any two odd numbers multiplied together produce this
result.

Sylvia.

Sylvia Else wrote:

>>> For any odd number x, there exists another odd number y, such that the
>>> product of x and y expressed in binary is a contiguous sequence of 1s,
>>> though I'll express them in hex. For example
>>> 3 * 1 = 3 5 * 3 = f 7 * 9 = 3f 9 * 7 = 3f
>> 11 * 93 = 3ff 13 * 315 = fff
>>> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>>
>> nonsense, 11 * 3 = 33 = 21h
>
> I didn't say that any two odd numbers multiplied together produce this
> result.

not sure I see the point, then. That must be true for all base numbers.

On 07-Nov-21 9:35 am, Gregor Bicha wrote:
> Sylvia Else wrote:
>
>>>> For any odd number x, there exists another odd number y, such that the
>>>> product of x and y expressed in binary is a contiguous sequence of 1s,
>>>> though I'll express them in hex. For example
>>>> 3 * 1 = 3 5 * 3 = f 7 * 9 = 3f 9 * 7 = 3f
>>> 11 * 93 = 3ff 13 * 315 = fff
>>>> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>>>
>>> nonsense, 11 * 3 = 33 = 21h
>>
>> I didn't say that any two odd numbers multiplied together produce this
>> result.
>
> not sure I see the point, then. That must be true for all base numbers.
>

Perhaps you're not familiar with David Fuller and his numerological posts.

Sylvia.

Sylvia Else wrote:

>>>> nonsense, 11 * 3 = 33 = 21h
>>>
>>> I didn't say that any two odd numbers multiplied together produce this
>>> result.
>>
>> not sure I see the point, then. That must be true for all base numbers.
>
> Perhaps you're not familiar with David Fuller and his numerological
> posts.

not sure that I am. In binary you only have 1s and 0s. So you could say
the same about the 0s, once you hit a binary power of 2, ie 2,4,8,16,32
etc, all the others are 0s.

On 07-Nov-21 9:56 am, Gregor Bicha wrote:
> Sylvia Else wrote:
>
>>>>> nonsense, 11 * 3 = 33 = 21h
>>>>
>>>> I didn't say that any two odd numbers multiplied together produce this
>>>> result.
>>>
>>> not sure I see the point, then. That must be true for all base numbers.
>>
>> Perhaps you're not familiar with David Fuller and his numerological
>> posts.
>
> not sure that I am. In binary you only have 1s and 0s. So you could say
> the same about the 0s, once you hit a binary power of 2, ie 2,4,8,16,32
> etc, all the others are 0s.
>

Maybe we're at cross purposes.

The product must equal 2^n - 1, for some n.

Sylvia.

On Friday, November 5, 2021 at 11:00:39 PM UTC-4, Sylvia Else wrote:
> For any odd number x, there exists another odd number y, such that the
> product of x and y expressed in binary is a contiguous sequence of 1s,
> though I'll express them in hex.
>
> For example
> 3 * 1 = 3
> 5 * 3 = f
> 7 * 9 = 3f
> ...
> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF

It is somewhat trivial to show, or to find a counterexample. I won't say which since its a prob for Fuller or other cranks.
You probably used the "formula" to find 816785180559426160758185055 [its a # I recognized] but note that that formula (which I will not divulge, yet?), is not minimal.

For instance, for 7 the formula works and gives 9. But 1 also produces the result: 7*1 = 111.
Better yet, for 97, the formula works too but doesn't give the smallest; the smallest is 2901803883615. Indeed,
97 * 2901803883615 = FFFFFFFFFFFF

Sylvia Else wrote:

>> not sure that I am. In binary you only have 1s and 0s. So you could say
>> the same about the 0s, once you hit a binary power of 2, ie 2,4,8,16,32
>> etc, all the others are 0s.
>
> Maybe we're at cross purposes.
> The product must equal 2^n - 1, for some n.

maybe just you, 32 is 20h, 16 is 10h, 8 is 08h etc. All others are ZEROS.
Use a brain.

On Saturday, November 6, 2021 at 7:39:07 PM UTC-4, Gregor Bicha wrote:

> maybe just you, 32 is 20h, 16 is 10h, 8 is 08h etc. All others are ZEROS.
> Use a brain.

DO NOT STROKE THE TROOLS.

Spam reported.
I incite others to do the same.

rotchm wrote:

> On Saturday, November 6, 2021 at 7:39:07 PM UTC-4, Gregor Bicha wrote:
>
>> maybe just you, 32 is 20h, 16 is 10h, 8 is 08h etc. All others are
>> ZEROS. Use a brain.
>
> DO NOT STROKE THE TROOLS. Spam reported.
> I incite others to do the same.

another imbecile not familiar with the powers of 2. Get the fuck out of
here, you stupid uneducated son of a bitch.

On 07-Nov-21 10:33 am, rotchm wrote:
> On Friday, November 5, 2021 at 11:00:39 PM UTC-4, Sylvia Else wrote:
>> For any odd number x, there exists another odd number y, such that the
>> product of x and y expressed in binary is a contiguous sequence of 1s,
>> though I'll express them in hex.
>>
>> For example
>> 3 * 1 = 3
>> 5 * 3 = f
>> 7 * 9 = 3f
>> ...
>> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>
> It is somewhat trivial to show, or to find a counterexample. I won't say which since its a prob for Fuller or other cranks.
> You probably used the "formula" to find 816785180559426160758185055 [its a # I recognized] but note that that formula (which I will not divulge, yet?), is not minimal.
>
> For instance, for 7 the formula works and gives 9. But 1 also produces the result: 7*1 = 111.
> Better yet, for 97, the formula works too but doesn't give the smallest; the smallest is 2901803883615. Indeed,
> 97 * 2901803883615 = FFFFFFFFFFFF
>

What I'm hoping[*] for is that David will find a method of determining y
for arbitrarily large x. Then I'll steal the method and use it to get
famous in an apparently unrelated field.

Sylvia.

[*] Hoping might be stretch ;)

Sylvia Else wrote:

> On 07-Nov-21 10:33 am, rotchm wrote:
>> For instance, for 7 the formula works and gives 9. But 1 also produces
>> the result: 7*1 = 111.
>> Better yet, for 97, the formula works too but doesn't give the
>> smallest; the smallest is 2901803883615. Indeed,
>> 97 * 2901803883615 = FFFFFFFFFFFF
>
> What I'm hoping[*] for is that David will find a method of determining y
> for arbitrarily large x. Then I'll steal the method and use it to get
> famous in an apparently unrelated field.

lol, but this rotchm is the most stupid uneducated sack of shit
disturbing with nonsense this group.

but the problem is simple, since in binary you only have 1s and 0s, the
same is applicable to 0s, as is it to 1s.

On 07-Nov-21 11:03 am, Gregor Bicha wrote:
> Sylvia Else wrote:
>
>> On 07-Nov-21 10:33 am, rotchm wrote:
>>> For instance, for 7 the formula works and gives 9. But 1 also produces
>>> the result: 7*1 = 111.
>>> Better yet, for 97, the formula works too but doesn't give the
>>> smallest; the smallest is 2901803883615. Indeed,
>>> 97 * 2901803883615 = FFFFFFFFFFFF
>>
>> What I'm hoping[*] for is that David will find a method of determining y
>> for arbitrarily large x. Then I'll steal the method and use it to get
>> famous in an apparently unrelated field.
>
> lol, but this rotchm is the most stupid uneducated sack of shit
> disturbing with nonsense this group.
>
> but the problem is simple, since in binary you only have 1s and 0s, the
> same is applicable to 0s, as is it to 1s.
>

rotchm appears to have figured this out already. It's far from clear to
me that you even understand the proposition.

Sylvia.

On Saturday, November 6, 2021 at 8:39:36 PM UTC-4, Sylvia Else wrote:

> rotchm appears to have figured this out already. It's far from clear to
> me that you even understand the proposition.

Of course the troll does not understand the question. Trolls are idiots by nature. Therefore they cannot understand nor solve the simplest of math problems. This troll has proven that he is too...a troll. Best just to ignore it and report it.

On Saturday, November 6, 2021 at 12:00:39 AM UTC-3, Sylvia Else wrote:
> For any odd number x, there exists another odd number y, such that the
> product of x and y expressed in binary is a contiguous sequence of 1s,
> though I'll express them in hex.
>
> For example
> 3 * 1 = 3
> 5 * 3 = f
> 7 * 9 = 3f
> 9 * 7 = 3f
> 11 * 93 = 3ff
> 13 * 315 = fff
>
> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>
> Sylvia.

If a, b and c are integers > 0 or 1, this is the equation to solve:

(2a - 1).(2b - 1) = 2ᶜ - 1

x = (2a - 1) , x: 1, 3, 5, 7, 9, 11, .... ; a: 1, 2, 3, 4, 5, 6, 7, 8, ......
y = (2b - 1) , y: 1, 3, 5, 7, 9, 11, .... ; b: 1, 2, 3, 4, 5, 6, 7, 8, ......

----------------

4ab - 2a - 2b + 1 = 2ᶜ - 1

4ab - 2a - 2b + 2 = 2ᶜ

This is an equation with two independent variables (a and b) and a linked parameter c, which should be c = c(a,b).

If there are solutions, it could be an infinite number of them. To test it, fix a with an arbitrary number for a (I choose prime numbers):

*** a = 3 gives: 12b - 6 - 2b + 2 = 2ᶜ

10b - 4 = 2ᶜ

c. log₂ 2 = log₂ (10b - 4)

If p = log₂ n, then 2ᵖ = n. For this, log₂ 2 = 1 and

c = log₂ (10b - 4) , c: 1, 2, 3, 4, 5, 6, 7, 8, .....

only find a solution for c = 4 and b = 2. The rest of values of c are irrational numbers from an Excel equation y = 1.6049 ln(x) + 2.854

*** Trying another value for b, choosing a = 37 gives

4.37.b - 2.37 - 2b + 2 = 2ᶜ
146 b - 72 = 2ᶜ

c = log₂ (146 b - 72) , c: 1, 2, 3, 4, 5, 6, 7, 8, .....

only find a solution for c = 9 and b = 4. The rest of values of c are irrational numbers from an Excel equation y = 1,6551 ln(x) + 6,5807

It seems that to solve (a, b, c), a not analytical method to search and discard is required, using a small program. But two variables
in a function c(a,b) makes impossible (IMHO) to obtain a general solution. Maybe relating b to a could provide an equation, but
the search for values of c that are integer require a program, anyway.

Hope this can help.

On 11/6/2021 8:40 PM, Sylvia Else wrote:
> On 07-Nov-21 11:03 am, Gregor Bicha wrote:

>> lol, but this rotchm is the most stupid uneducated sack of shit
>> disturbing with nonsense this group.
>>
>> but the problem is simple, since in binary you only have 1s and 0s, the
>> same is applicable to 0s, as is it to 1s.
>>
>
> rotchm appears to have figured this out already. It's far from clear to
> me that you even understand the proposition.

Again, that's the nymshifting troll, whose only purpose is to generate
noise in here. It deliberately "misunderstood" the problem to get
people squabbling about what an idiot it is.

On 07-Nov-21 3:15 pm, Richard Hertz wrote:
> On Saturday, November 6, 2021 at 12:00:39 AM UTC-3, Sylvia Else wrote:
>> For any odd number x, there exists another odd number y, such that the
>> product of x and y expressed in binary is a contiguous sequence of 1s,
>> though I'll express them in hex.
>>
>> For example
>> 3 * 1 = 3
>> 5 * 3 = f
>> 7 * 9 = 3f
>> 9 * 7 = 3f
>> 11 * 93 = 3ff
>> 13 * 315 = fff
>>
>> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>>
>> Sylvia.
>
>
> If a, b and c are integers > 0 or 1, this is the equation to solve:
>
> (2a - 1).(2b - 1) = 2ᶜ - 1
>
> x = (2a - 1) , x: 1, 3, 5, 7, 9, 11, .... ; a: 1, 2, 3, 4, 5, 6, 7, 8, .....
> y = (2b - 1) , y: 1, 3, 5, 7, 9, 11, .... ; b: 1, 2, 3, 4, 5, 6, 7, 8, .....
>
> ----------------
>
> 4ab - 2a - 2b + 1 = 2ᶜ - 1
>
> 4ab - 2a - 2b + 2 = 2ᶜ
>
> This is an equation with two independent variables (a and b) and a linked parameter c, which should be c = c(a,b).
>
> If there are solutions, it could be an infinite number of them. To test it, fix a with an arbitrary number for a (I choose prime numbers):
>
> *** a = 3 gives: 12b - 6 - 2b + 2 = 2ᶜ
>
> 10b - 4 = 2ᶜ
>
> c. log₂ 2 = log₂ (10b - 4)
>
> If p = log₂ n, then 2ᵖ = n. For this, log₂ 2 = 1 and
>
> c = log₂ (10b - 4) , c: 1, 2, 3, 4, 5, 6, 7, 8, .....
>
> only find a solution for c = 4 and b = 2. The rest of values of c are irrational numbers from an Excel equation y = 1.6049 ln(x) + 2.854
>
> *** Trying another value for b, choosing a = 37 gives
>
> 4.37.b - 2.37 - 2b + 2 = 2ᶜ
> 146 b - 72 = 2ᶜ
>
> c = log₂ (146 b - 72) , c: 1, 2, 3, 4, 5, 6, 7, 8, .....
>
> only find a solution for c = 9 and b = 4. The rest of values of c are irrational numbers from an Excel equation y = 1,6551 ln(x) + 6,5807
>
> It seems that to solve (a, b, c), a not analytical method to search and discard is required, using a small program. But two variables
> in a function c(a,b) makes impossible (IMHO) to obtain a general solution. Maybe relating b to a could provide an equation, but
> the search for values of c that are integer require a program, anyway.
>
> Hope this can help.

Ye Gods!

Perhaps you should stick to attacking relativity.

As rotchm has already noted, there is a simple analytical way of
obtaining a value for y for a given x.

In any case, if one is resorting to using a computer, the obvious course
is a systematic search by trial dividing sucessive values of 2^n - 1 by x.

Sylvia.

On Sunday, November 7, 2021 at 2:36:47 AM UTC-3, Sylvia Else wrote:

<snip>

> As rotchm has already noted, there is a simple analytical way of
> obtaining a value for y for a given x.
>
> In any case, if one is resorting to using a computer, the obvious course
> is a systematic search by trial dividing sucessive values of 2^n - 1 by x..

I was thinking that the equation

y = (2b - 1) = (2ᶜ - 1)/(2a - 1)

could be explored as a matrix in the memory with:

- a range of values for c between 1 and N (you choose the limit, depending on the available memory). They are in a ROW.

- under each row, you expand colums for value of a between 1 and M (memory again) containing (2ᶜ - 1)/(2a - 1) values
on each cell of the N x M matrix. You keep a column for values of a for separate.

- now, if you launch a program to find cells with integers ONLY (up to the 5th.decimal, for sure), the program will deliver
matches that render couples of (a, c) values exactly (which is a double column).

- now, it only rest to solve y = (2b - 1) = (2ᶜ - 1)/(2a - 1) to obtain exact values (a, b, c). You then derive x and y from them.

Glad to contribute with this interesting puzzle.

On Sunday, November 7, 2021 at 12:15:24 AM UTC-4, Richard Hertz wrote:

> If a, b and c are integers > 0 or 1, this is the equation to solve:
>
> (2a - 1).(2b - 1) = 2ᶜ - 1

Using 2b+1 instead would simplify a little. No more "-1" to tag along.

> x = (2a - 1) , x: 1, 3, 5, 7, 9, 11, .... ; a: 1, 2, 3, 4, 5, 6, 7, 8, ......
> y = (2b - 1) , y: 1, 3, 5, 7, 9, 11, .... ; b: 1, 2, 3, 4, 5, 6, 7, 8, ......

> 4ab - 2a - 2b + 1 = 2ᶜ - 1
> 4ab - 2a - 2b + 2 = 2ᶜ
> This is an equation with two independent variables (a and b) and a linked parameter c, which should be c = c(a,b).
> If there are solutions, it could be an infinite number of them. To test it, fix a with an arbitrary number for a (I choose prime numbers):
>
> *** a = 3 gives: 12b - 6 - 2b + 2 = 2ᶜ
>...
> c = log₂ (10b - 4) , c: 1, 2, 3, 4, 5, 6, 7, 8, .....
>
> only find a solution for c = 4 and b = 2. The rest of values of c are irrational numbers from an Excel equation y = 1.6049 ln(x) + 2.854

Eeech... You havent tried hard havent you!
With c = 8, what do you get? [do it by hand, its more precise than Excel]..
Now, you see whats going on? See the pattern? Can you explain it now?
What are thus the other solutions?

Now, can you generalize to any "a" ? Yes/no/why not/counterexample ?

And the question of interest is:

For any [odd] M, is there an [odd] N such that M*N = 11...1_₂?
Yes or no?

On Friday, November 5, 2021 at 8:00:39 PM UTC-7, Sylvia Else wrote:
> For any odd number x, there exists another odd number y, such that the
> product of x and y expressed in binary is a contiguous sequence of 1s,
> though I'll express them in hex.
>
> For example
> 3 * 1 = 3
> 5 * 3 = f
> 7 * 9 = 3f
> 9 * 7 = 3f
> 11 * 93 = 3ff
> 13 * 315 = fff
>
> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
>
> Sylvia.

b=(2^(c-1)+a-1)/(2a-1)

The above is a diophantic equation that admits integer solution only if a=EVEN. This observation halves the number of searches.

By symmetry:

a=(2^(c-1)+b-1)/(2b-1)

The above is a diophantic equation that admits integer solution only if b=EVEN.

On Saturday, November 6, 2021 at 4:59:37 PM UTC-7, Sylvia Else wrote:
> On 07-Nov-21 10:33 am, rotchm wrote:
> > On Friday, November 5, 2021 at 11:00:39 PM UTC-4, Sylvia Else wrote:
> >> For any odd number x, there exists another odd number y, such that the
> >> product of x and y expressed in binary is a contiguous sequence of 1s,
> >> though I'll express them in hex.
> >>
> >> For example
> >> 3 * 1 = 3
> >> 5 * 3 = f
> >> 7 * 9 = 3f
> >> ...
> >> 97 * 816785180559426160758185055 = FFFFFFFFFFFFFFFFFFFFFFFF
> >
> > It is somewhat trivial to show, or to find a counterexample. I won't say which since its a prob for Fuller or other cranks.
> > You probably used the "formula" to find 816785180559426160758185055 [its a # I recognized] but note that that formula (which I will not divulge, yet?), is not minimal.
> >
> > For instance, for 7 the formula works and gives 9. But 1 also produces the result: 7*1 = 111.
> > Better yet, for 97, the formula works too but doesn't give the smallest; the smallest is 2901803883615. Indeed,
> > 97 * 2901803883615 = FFFFFFFFFFFF
> >
> What I'm hoping[*] for is that David will find a method of determining y
> for arbitrarily large x. Then I'll steal the method and use it to get
> famous in an apparently unrelated field.
>
> Sylvia.
>
> [*] Hoping might be stretch ;)

Dave "Kronos Prime" Fuller works out those fractions mostly from
the size of scale factors and trigonometry.

Otherwise though it's a matter of factors what some binary rep-unit,
which is simply enough "one less than on the order of", making for
numbers what in some arbitrarily larger order, they're a small and
large number +-1 for often something usual about "x^2 -1 = (x+1)(x-1)".

One of the numerical techniques like Fuller's that makes for "abitrarily
precise rationication", is it's usual notion that "such large constants
have their order erased as significant digits in usual products", instead
he's so scaffolded up that it results the orders of scale are so large as
that their measurements the one way and the other, see this "ratiocination"
fall out while still landing at either end on the constants.

I.e. that the measurements are so precise either way make for a
large enough or precise constant, that the current order of experiment
defines how long David "Kronos Prime" Fuller rep-value constants like
..777... fit, then for where they arrive from as under geometric series
or scale factors themselves.

On Sunday, November 7, 2021 at 10:17:09 AM UTC-5, Dono. wrote:

> b=(2^(c-1)+a-1)/(2a-1)
>
> The above is a diophantic equation that admits integer solution only if a=EVEN. This observation halves the number of searches.

Let c = 6 and a = 5 [which is an odd integer].
b=(2^(c-1)+a-1)/(2a-1) = 4 [an integer].

Thus, there are solutions when "a" is odd.

On Saturday, November 6, 2021 at 11:12:51 PM UTC-7, Richard Hertz wrote:
> On Sunday, November 7, 2021 at 2:36:47 AM UTC-3, Sylvia Else wrote:
> > For any odd number x, there exists another odd number y, such that the
> > product of x and y expressed in binary is a contiguous sequence of 1s.
>
> ... now, if you launch a program to find cells with integers ONLY (up to the
> 5th.decimal, for sure), the program will deliver ...

This thread sheds some light on a recent discussion in another thread about the difference between scientists and mathematicians on one hand, and engineers on the other. An engineer might try to use a "brute force" method of trial and error on a computer, searching for an odd integer y for a given odd x such that xy is a binary repunit. This approach has some obvious disadvantages: First, it's very laborious and inefficient; second, it can never prove that there is such a y for every odd x; and third, it gives no real insight into the solutions.

Now, an amateur mathematician (nearly 400 years ago) found the simple theorem that gives the explicit answer, and another mathematician (about 300 years ago) refined that theorem to give the minimum y value explicitly for any x, and they did this without any electronic computers at all. Moreover, the theorem they proved is used very extensively in modern technology for things like digital data error detection and corrections codes, and so on. Of course, many of the engineers who use CRCs are not experts in how or why they work. They rely on the mathematicians.

On Sunday, November 7, 2021 at 7:46:08 AM UTC-8, rotchm wrote:
> On Sunday, November 7, 2021 at 10:17:09 AM UTC-5, Dono. wrote:
>
> > b=(2^(c-1)+a-1)/(2a-1)
> >
> > The above is a diophantic equation that admits integer solution only if a=EVEN. This observation halves the number of searches.
> Let c = 6 and a = 5 [which is an odd integer].
> b=(2^(c-1)+a-1)/(2a-1) = 4 [an integer].
>
> Thus, there are solutions when "a" is odd.
you are right