Deeper than primes

epix · Oct 3, 2010

doronshadmi said:
Cardinality is exactly a measurement unit of the size (or magnitude) of existence.

"In mathematics, the cardinality of a set is a measure of the "number of elements of the set". For example, the set A = {2, 4, 6} contains 3 elements, and therefore A has a cardinality of 3."

"Really? I didn't know that. So like {G, o, d} = 3?"

doronshadmi · Oct 3, 2010

The Man said:
Technically and more specifically it is a “1-dim element” defined by (at least) two “0-dim elements”.

Again a line segment (a “1-dim element”) is defined by two points (“0-dim elements”), the absurdities are again and remain entirely yours.

A line segment is a complex result of the linkage of at least 1-dim element and two 0-dim elements.

Again, you do not undrstand that only 0-dim elements can't define a size > 0.

epix · Oct 3, 2010

doronshadmi said:
Just take to account that that S = 2(a+b+c+d+...) does not have a sum.

We'll get to it. The a, b, c, d, ... line segments are defined by the position of the baselines in the triangle, so we'll see how they look like, despite your claim that a+b doesn't have a sum. If 1+2 have a sum (like 3?) then a+b should have a sum as well.

jsfisher · Oct 3, 2010

doronshadmi said:
A line segment is a complex result of the linkage of at least 1-dim element and two 0-dim elements.

Again, you do not undrstand that only 0-dim elements can't define a size > 0.

Again, you do not understand that stuff you simply make up isn't necessarily true.

The Man · Oct 3, 2010

epix said:
"In mathematics, the cardinality of a set is a measure of the "number of elements of the set". For example, the set A = {2, 4, 6} contains 3 elements, and therefore A has a cardinality of 3."

"Really? I didn't know that. So like {G, o, d} = 3?"

Well a bird (or cardinal) in the hand is worth two in the bush. So since “OM” is “not less than” Hand/Bush “interaction”, giving Doron not just ‘the bird’ but three ‘magnitudes of existence’ in units of ‘the bird’.

doronshadmi · Oct 3, 2010

epix said:
"In mathematics, the cardinality of a set is a measure of the "number of elements of the set". For example, the set A = {2, 4, 6} contains 3 elements, and therefore A has a cardinality of 3."

"Really? I didn't know that. So like {G, o, d} = 3?"

I do not care about the traditional notion of Cardinality because by OM it is extended to measure directly Emptiness (that has no predecessor) Fullness (that has nor successor), and any level between them (that has predecessor AND successor).

doronshadmi · Oct 3, 2010

epix said:
We'll get to it. The a, b, c, d, ... line segments are defined by the position of the baselines in the triangle, so we'll see how they look like, despite your claim that a+b doesn't have a sum. If 1+2 have a sum (like 3?) then a+b should have a sum as well.

Finitly many positive values have a sum.

Infinitly many positive values do not have a sum.

The Man · Oct 3, 2010

doronshadmi said:
A line segment is a complex result of the linkage of at least 1-dim element and two 0-dim elements.

Again you simply don’t understand that the “linkage” is specifically a line segment being defined by points and that “linkage” is evidently only “complex” for you.

doronshadmi said:
Again, you do not undrstand that only 0-dim elements can't define a size > 0.

Again you, apparently intentionally, do understand that you simply claiming something does not make it a fact and that points can and do define a line segment is a fact.

doronshadmi · Oct 3, 2010

jsfisher said:
Again, you do not understand that stuff you simply make up isn't necessarily true.

You are invided to provide the proof that an addition of infinitely many 0 sizes is resulted by a size > 0.

doronshadmi · Oct 3, 2010

The Man said:
Again you simply don’t understand that the “linkage” is specifically a line segment being defined by points and that “linkage” is evidently only “complex” for you.

Call it in any name you like, it does not change the fact that any amount 0-dim elements can't be a 1-dim element.

By avoid what you call "linkage", please define two 0-dim elements with different values.

jsfisher · Oct 3, 2010

doronshadmi said:
Call it in any name you like, it does not change the fact that any amount 0-dim elements can't be a 1-dim element.

And yet, Mathematics bends to neither your whim nor your misunderstanding, and lines and line segments continue to consist of nothing but points.

By avoid what you call "linkage", please define two 0-dim elements with different values.

Oh, come on, Doron. Now you are just being silly. The simple mechanism of taking a point's location as a value is sufficient.

The Man · Oct 3, 2010

doronshadmi said:
Call it in any name you like, it does not change the fact that any amount 0-dim elements can't be a 1-dim element.

~~What name do you think I called you?~~

Your assertions do not change the fact that two points define a line segment.

doronshadmi said:
By avoid what you call "linkage", please define two 0-dim elements with different values.

Well, since you’re the one claiming points don’t define a line segment, you’ll have to do that.

doronshadmi · Oct 3, 2010

jsfisher said:
And yet, Mathematics bends to neither your whim nor your misunderstanding, and lines and line segments continue to consist of nothing but points.

Here we go, by jsfisher limited reasoning 1-dim does not exist unless there is a collection of 0-dim elements.

jsfisher said:
Oh, come on, Doron. Now you are just being silly. The simple mechanism of taking a point's location as a value is sufficient.

It is simple if a 1-dim element is also used by this mechanism.

doronshadmi · Oct 3, 2010

The Man said:
~~What name do you think I called you?~~

Your assertions do not change the fact that two points define a line segment.

Well, since you’re the one claiming points don’t define a line segment, you’ll have to do that.

Your definition do not change the fact that at least 1-dim element AND more than one 0-dim element, are involved.

epix · Oct 3, 2010

doronshadmi said:
Finitly many positive values have a sum.

Infinitly many positive values do not have a sum.

If the sum of the positive values diverges then the value of the sum approaches infinity; if the sum converges, then it's value doesn't exceed a certain number called "limit." That limit is considered the value of the summation. It works that way in calculus where integration is basically a summation.

jsfisher · Oct 3, 2010

doronshadmi said:
Here we go, by jsfisher limited reasoning 1-dim does not exist unless there is a collection of 0-dim elements.

If you wish to develop an alternate geometry without points, knock yourself out. I bet you can't, though. You have been unable to define even the simplest of your made-up vocabulary, so it would be impossible for you to build anything more complicated.

It is simple if a 1-dim is also used by this mechanism.

So what? That wasn't the challenge. Can't you follow even your own conversation?

The 5 year olds are giggling loudly now, Doron. Please stop. They really need to focus on their lessons.

epix · Oct 3, 2010

jsfisher said:
If you wish to develop an alternate geometry without points, knock yourself out.

You know, that Organic Mathematics reminds me Topology -- the other Doron's pointless attempt to modernize math. It just doesn't go anywhere, as many have already found out

epix · Oct 3, 2010

The Man said:
~~What name do you think I called you?~~

Your assertions do not change the fact that two points define a line segment.

What name? You know very well what name. Don't make me to remind you.

In mathematics, point-free geometry is a geometry whose primitive ontological notion is region rather than point.

What a ride. LOL.

doronshadmi · Oct 3, 2010

jsfisher said:
If you wish to develop an alternate geometry without points, knock yourself out.

You still don't get it isn't it jsfisher?

We are not talking here about Geometry, but about Locality and Non-locality as fundamental qualities of any given mathematical branch.

A point is nothing but the minimal geometrical aspect of Locality, and a line is nothing but the minimal geometrical aspect of Non-locality.

By following this fundamental notion a line is not a collection of points, a plane is not a collection of points or lines, a volume is not a collection of planes, lines or points, etc … ad infinitum.

If linked they are resulted by complexities.

jsfisher · Oct 3, 2010

doronshadmi said:
You still don't get it isn't it jsfisher?

I get far more than you can comprehend.

We are not talking here about Geometry, but about Locality and Non-locality as fundamental qualities of any given mathematical branch.

And, yet, you cannot define any of your terminology. The best you can do is misuse established terms.

A point is nothing but the minimal geometrical aspect of Locality, and a line is nothing but the minimal geometrical aspect of Non-locality.

It is amusing to me that, according to you, you are not talking about geometry, but nevertheless here you are talking about geometry.

By following this fundamental notion a line is not a collection of points, a plane is not a collection of points or lines, a volume is not a collection of planes, lines or points, etc … ad infinitum.

Nonsense. You have failed completely to distinguish your line segments from the normal ones. Points do a splendid job of covering your line segments, despite your contradictions and protestations claiming it isn't so.

doronshadmi · Oct 3, 2010

jsfisher said:
I get far more than you can comprehend.

Not even in your dreams.

By ignoring your ignorance let's continue to develop OM.

By using the notion of dimensional spaces, we notate these two extreme cases:

In Case A, the dimensional spaces are not resulted by complexities:

∞(
0(),1(),2(),3(), …
)

In Case B, the dimensional spaces are resulted by complexities:

∞(
… 3(2(1(0()))) …
)

There are infinitely many cases between cases A and B.

doronshadmi · Oct 3, 2010

epix said:
You know, that Organic Mathematics reminds me Topology -- the other Doron's pointless attempt to modernize math. It just doesn't go anywhere, as many have already found out

epix, Pointless_topology is also under Non-locality\Locality Linkage, because whenever multiplicity is used, there is Non-locality among the considered elements. so?

The Man · Oct 3, 2010

doronshadmi said:
Your definition do not change the fact that at least 1-dim element AND more than one 0-dim element, are involved.

“do not change the fact”? Doron, the definition of a line segment establishes those facts.

The Man · Oct 3, 2010

doronshadmi said:
Not even in your dreams.

By ignoring your ignorance let's continue to develop OM.

By using the notion of dimensional spaces, we notate these two extreme cases:

In Case A, the dimensional spaces are not resulted by complexities:

∞(
0(),1(),2(),3(), …
)

In Case B, the dimensional spaces are resulted by complexities:

∞(
… 3(2(1(0()))) …
)

There are infinitely many cases between cases A and B.

No Doron, thinking that you are in any way developing your “OM” would simply be ignoring your own ignorance.

epix · Oct 3, 2010

doronshadmi said:
Just take to account that that S = 2(a+b+c+d+...) does not have a sum.

Well, it does. I would draw a pic, but I think that would be a waste of time. I translate the rendition of the sum from OM to the normal language, while you get you drawing.

The terms of the sum are

a₀, a₁, a₃, ... , a_n-1, a_n

where n -> ∞

The term a_j has an algebraic equivalent in

a_j = (3^j*a_j)/(2^2j+3)

and the sum of the terms is

∞
∑ (3^j*a_j)/(2^2j+3) = a/2
j=0

where the 'a' without index is the length of the side of the triangle. That means the sum equal 1/2 of the side. And that's the proof that the combined length of the segment of each iteration keeps constant and equals the side of the triangle.

This conclusion will surely not agree with the OM conclusion, which uses different tools.

Well, the universe uses the standard ways to prove stuff, but I don't see how this gets us any closer to the Organic Number #18.

jsfisher · Oct 3, 2010

epix said:
Well, it does. I would draw a pic, but I think that would be a waste of time. I translate the rendition of the sum from OM to the normal language, while you get you drawing.

He already drew a picture that makes the point.

It should be obvious that you don't need to consider each individual Koch curve generation. At each iteration, S is simply the length of the inverted triangle's base (which is X, although Doron never says so) minus the width of the current generation. You don't need the 2(a+b+c+...) to tell you how S got to where it is; it just need to know where it is.

Then, since the width gets vanishingly small with each iteration (i.e. approaches 0 in the limit), S approaches the full length of the base.

Ironic that Doron's so-called "proof without words" demonstrates the opposite of what is supposes.

epix · Oct 4, 2010

jsfisher said:
He already drew a picture that makes the point.

It should be obvious that you don't need to consider each individual Koch curve generation. At each iteration, S is simply the length of the inverted triangle's base (which is X, although Doron never says so) minus the width of the current generation. You don't need the 2(a+b+c+...) to tell you how S got to where it is; it just need to know where it is.

Then, since the width gets vanishingly small with each iteration (i.e. approaches 0 in the limit), S approaches the full length of the base.

Ironic that Doron's so-called "proof without words" demonstrates the opposite of what is supposes.

No, you don't really need the sum the way Doron showed. The sum of the position of the baselines of each iteration will do -- it's more straightforward and frees the clutter in the summand. I'll do it once again the easier way.

I made a peculiar mistake here

a_j = (3^j*a_j/(2^2j+3)

a_j appears in both side of the equation. Also, the term can be further reduced, and so when s₀ is the length of the side of the triangle, then

a_j = s₀*(3/4)^j/8

The sum is then

∞
∑ s₀*(3/4)^j/8 = s₀/2
j=0

I don't think that it matters that much, coz Doron will junk it all -- mistake or no mistake -- when Fullness and Emptiness carry him back to the Battle of Reason. LOL. That's why I don't want to do this step-by-step. Doron has already chosen his destiny and rolls over anything that's "non-local" to OM.

doronshadmi · Oct 4, 2010

jsfisher said:
S approaches the full length of the base.

I agree, S only approaches constant X > 0, such that S < constant X > 0 by fog 0.000...3/4

jsfisher does not distinguish between "reaches" and "approaches".

jsfisher said:
You don't need the 2(a+b+c+...) to tell you how S got to where it is; it just need to know where it is.

You can't, beccause S is an infinite interpolation that does not have a strict value.

jsfisher, you simply can't grasp the irresistible fact that 1-dim space is irreducible to 0-dim space, exactly because no given k-dim space (where k=0 to ∞) is defined by any other dim space, ad infinitum, where this notion is notated by:

∞(
0(),1(),2(),3(), …
)

doronshadmi · Oct 4, 2010

The Man said:
“do not change the fact”? Doron, the definition of a line segment establishes those facts.

No The Man, the definition of a line segment is possible exactly becuse two spaces (0-dim and 1-dim) that are not derive from each other, define a complex, called "line segment".

The Man, you simply can't grasp the irresistible fact that 1-dim space is irreducible to 0-dim space, exactly because no given k-dim space (where k=0 to ∞) is defined by any other dim space, ad infinitum, where this notion is notated by:

∞(
0(),1(),2(),3(), …
)

epix · Oct 4, 2010

doronshadmi said:
You can't, beccause S is an infinite interpolation that does not have a strict value.

You need to redefine the word "interpolation" so its meaning wouldn't contradict conclusions that OM makes.

In the mathematical subfield of numerical analysis, interpolation is a method of constructing new data points within the range of a discrete set of known data points.

Interpolation allows to create new points within a given range; it allows gaps to be filled with points whose number may approach infinity, if it's needed. But that's the idea that is a big no-no in OM, coz a finite line segment cannot be comprised of points whose number approaches infinity, right?

Once you redefine the interpolation process to fit OM needs, you should change the word as well -- you can call it "Nebe" after the dude who was a head of Interpol during WW2, so Emptiness and Fullness got someone to play with.

Btw, the Koch formations don't live in Number Theory which recognizes only "strict" values. In other words Koch fractals are not a case of discrete math, so why are you treating it like that?

doronshadmi · Oct 5, 2010

epix said:
Btw, the Koch formations don't live in Number Theory which recognizes only "strict" values. In other words Koch fractals are not a case of discrete math, so why are you treating it like that?

You still miss it.

The notion that between any arbitrary given pair of points there is at least another point, is a permanent fact that is not changed, no matter how many scale levels are considered.

The continuum is at least 1-dim space, and no amount of 0-dim spaces along it can be a 1-dim space.

n = 1 to ∞

The same fact holds among any n-1 dim space and n dim space upon infinitely many dimesional spaces.

jsfisher · Oct 5, 2010

doronshadmi said:
You still miss it.

The notion that between any arbitrary given pair of points there is at least another point, is a permanent fact that is not changed, no matter how many scale levels are considered.

The continuum is at least 1-dim space, and no amount of 0-dim spaces along it can be a 1-dim space.

n = 1 to ∞

The same fact holds among any n-1 dim space and n dim space upon infinitely many dimesional spaces.

Saying it over and over doesn't make it true, Doron. The fact remains that in real Mathematics points completely cover a line.

You may want doronetics to behave differently. Go for it!! Please be so kind, though, to (1) make it clear to everyone you are not talking about real Mathematics, just doronetics, and (2) identify a location on one of your doronetics lines where there is no point.

epix · Oct 5, 2010

doronshadmi said:
By ignoring your ignorance let's continue to develop OM.

By using the notion of dimensional spaces, we notate these two extreme cases:

In Case A, the dimensional spaces are not resulted by complexities:

∞(
0(),1(),2(),3(), …
)

In Case B, the dimensional spaces are resulted by complexities:

∞(
… 3(2(1(0()))) …
)

There are infinitely many cases between cases A and B.

You are trying to apply your "axiomatic framework" to the fractals, but the situation is more complicated, coz the space where they live may not be defined by a dimension that is an integer alone.

For many shapes that are often considered in mathematics, physics and other disciplines, the Hausdorff dimension is an integer. However, sets with non-integer Hausdorff dimension are important and prevalent. Benoît Mandelbrot, a popularizer of fractals, advocates that most shapes found in nature are fractals with non-integer dimension, explaining that "clouds are not spheres, mountains are not cones, coastlines are not circles, and bark is not smooth, nor does lightning travel in a straight line."

Also remember that a fractal point can be multi-dimensional (color) where its dimensionality is given by a number of iterations.
http://www2.warwick.ac.uk/fac/sci/sbdtc/students/2007/robert_gardner/the_past/mandelbrot_set.gif

The Man · Oct 6, 2010

doronshadmi said:
No The Man, the definition of a line segment is possible exactly becuse two spaces (0-dim and 1-dim) that are not derive from each other, define a complex, called "line segment".

Again Doron, the definition of a line segment explicitly derives from the definition of a point. Again this simple fact only seems “complex” to you.

doronshadmi said:
The Man, you simply can't grasp the irresistible fact that 1-dim space is irreducible to 0-dim space, exactly because no given k-dim space (where k=0 to ∞) is defined by any other dim space, ad infinitum, …

Again Doron “you simply can't grasp the irresistible fact that” a line segment is defined by at least two points.

doronshadmi said:
where this notion is notated by:

∞(
0(),1(),2(),3(), …
)

You can notate you notions by any means that suit you, though please let us know when you can acctualy notate any self-consistency, general consistency or practical application (like an ‘infinite interpolation energy’ toaster) of your notions.

epix · Oct 6, 2010

doronshadmi said:
Just take to account that that S = 2(a+b+c+d+...) does not have a sum.

You wanted to say that the addition of infinitely many positive values doesn't have a sum, right? Well, it kind of make sense, coz the process of addition never stops and therefore the sum changes with each added number. In that case, how can anyone compare the ever-changing result with another number, such as a constant?

You are living under the impression that there is no sum, and that’s because you would define the task of adding the series of positive integers this way

S = 1 + 2 + 3 + 4 + ...

as it can be seen from the way you did it in S = 2(a + b + c + d + ...)

That addition really doesn't have any sum, coz one important element is missing. The instruction to sum the series of positive integers must look like this

S = 1 + 2 + 3 + 4 + ... + n-1 + n

where n → ∞.

The number ‘n’ is important, coz it is defined: n = "a value that approaches infinity."

You may disagree and point out that the definition of 'n' is useless, coz it would take an infinite amount of time to complete the addition -- the process would never end, and therefore "there is no sum."

Actually even the task of adding the first hundred numbers could make someone really impatient . . .

Another famous story has it that in primary school after the young Gauss misbehaved, his teacher, J.G. Büttner, gave him a task add a list of integers in arithmetic progression; as the story is most often told, these were the numbers from 1 to 100. The young Gauss reputedly produced the correct answer within seconds, to the astonishment of his teacher and his assistant Martin Bartels.

Gauss's presumed method was to realize that pairwise addition of terms from opposite ends of the list yielded identical intermediate sums1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on, for a total sum of 50 × 101 = 5050.

And so, due to little Carl and others before him, the infinite addition of the series of whole numbers does have a sum:

S = 1 + 2 + 3 + 4 + ... + n-1 + n = (n² + n)/2.

So does any infinite summation.

When an infinite series converges, the sum includes ‘n’ as well and so the algebraic term can be compared with other terms that include ‘n’. But an arbitrary constant doesn’t include it, and so there is “strictly speaking” no way that the result of the summation would ever equal a constant. But remember that there are many ways to skin the cat in math. Here is an infinite convergent series whose sum is Pi.

∞
4∑ (-1^k)/(2k +1) = Pi
k=0

π (sometimes written pi) is a mathematical constant whose value is the ratio of any circle's circumference to its diameter in Euclidean space; this is the same value as the ratio of a circle's area to the square of its radius. It is approximately equal to 3.14159265 in the usual decimal notation.

Pi can be also expressed in an exact form, such as Pi = c/d (circle circumference/diameter). It all depends on the skill of the mathematician to convert a sum of a convergent series into an exact form and do the comparison with a constant. Sometimes it is possible, more often it is not – it all depends on the particular problem to be solved.

So saying that an infinite series doesn’t have a sum is not true, as any high school kid would confirm so. But they don’t teach OM in the schools, and in this field anything seems to be possible and at the same time impossible.

S = 1 + 2 + 3 + 9 + 24 + 76 + 236 + ... + n-1 + n. (n → ∞)

How does the summation formula look like, Doron?

DazzaD · Oct 7, 2010

epix said:
Pi can be also expressed in an exact form, such as Pi = c/d (circle circumference/diameter). It all depends on the skill of the mathematician to convert a sum of a convergent series into an exact form and do the comparison with a constant. Sometimes it is possible, more often it is not – it all depends on the particular problem to be solved.

So saying that an infinite series doesn’t have a sum is not true, as any high school kid would confirm so. But they don’t teach OM in the schools, and in this field anything seems to be possible and at the same time impossible.

And can I point out to Doron that

pi is not contingent upon "circumference/diameter", it has a clearly defined value in mathematics regardless of this application in geometry (something he may want to consider when invoking the Koch-type diagrams)..

and that A Level maths students in the UK (ages 16-18) are well versed in the "sum to infinity of a geometric series" and would struggle greatly to understand one line of his explanation about why that series does NOT 'have' a finite sum.

doronshadmi · Oct 8, 2010

jsfisher said:
Saying it over and over doesn't make it true, Doron. The fact remains that in real Mathematics points completely cover a line.

n = 1 to ∞

Saying it over and over doesn't make it true, jsfisher. The fact remains that in real Mathematics points are not completely cover a line, simply because no amount of 0 to n-1 dimensional spaces is n dimensional space ad infinitum ... exactly because any given dimensional space is not a collection of the previous dimensional spaces.

Furthermore, the strict identification of dimensional spaces of the form

A=0()
B=1()

∞(
A,B
)

is just a particular case of Frame (1,1) under X-Redundancy x Y-Uncertainty Distinction Tree, as follows:

Code:

(AB,AB) (AB,A)  (AB,B)  (AB)    (A,A)   (B,B)   (A,B)   (A)     (B)     ()

A * *   A * *   A * .   A * .   A * *   A . .   A * .   A * .   A . .   A . .
  | |     | |     | |     | |     | |     | |     | |     | |     | |     | |
B *_*   B *_.   B *_*   B *_.   B ._.   B *_*   B ._*   B ._.   B *_.   B ._.

(2,2) = (AB,AB)
(2,1) = (AB,A),(AB,B)
(2,0)=  (AB)
[COLOR="magenta"][B](1,1)[/B][/COLOR] = (A,A),(B,B), [COLOR="magenta"][B](A,B)[/B][/COLOR]
(1,0)=  (A),(B)
(0,0)=  ()

doronshadmi · Oct 8, 2010

jsfisher said:
identify a location on one of your doronetics lines where there is no point.

You are talking about the the complex form

∞(
1(0())
)

that does not change the fact about the non-complex form

∞(
0(),1()
)

The traditional reasoning can't distinguish between the complex and the non-complex.

doronshadmi · Oct 8, 2010

epix said:
You are trying to apply your "axiomatic framework" to the fractals, but the situation is more complicated, coz the space where they live may not be defined by a dimension that is an integer alone.

Wrong.

Cardinality is also a measurement unit of an infinite interpolation between any pair of strict values where (for example) the particular case of 1(0()) is equivalent to [0,1].

In 1(0()) the depth between 0() and 1() is considered, where in [0,1] no depth is considered (the system is flat).

epix said:
Also remember that a fractal point can be multi-dimensional (color) where its dimensionality is given by a number of iterations.
http://www2.warwick.ac.uk/fac/sci/sbdtc/students/2007/robert_gardner/the_past/mandelbrot_set.gif

What you call "a number of iterations" is equivalent to interpolation, which is strict if it is finite interpolation, or non-strict if it is infinite interpolation (in infinite interpolation, non-local numbers like 0.999…[base 10] and 0.000…1[base 10] are used).

doronshadmi · Oct 8, 2010

The Man said:
Again Doron, the definition of a line segment explicitly derives from the definition of a point.

Worng, a line segement is not less than 0() AND 1() ,which is the complexity 1(0()).

Deeper than primes

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