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In my previous post on DNA, I mentioned the following argument:

A) DNA is information.
B) Information cannot arise from a random, non-directed process.
C) Darwinism requires DNA to have arisen from a random, non-directed process.
D) Therefore, Darwinism cannot explain DNA.

In my first post, I demonstrated A) DNA is information. In this post, I will demonstrate B) Information cannot arise from a random, non-directed process.

The first thing to note is an example that Apolonio brought up. He said:

For example, we can conceive of a case where a person knocks over a scrabble box and the letters I Love You comes out with that order.

While this would be a semi-random process creating information, it is not using foundational forces. The specific example requires a person to knock over the Scrabble box. But even if we adjust for that and make it gravity pulling a box off a shelf or something similar, Scrabble tiles are not foundational in nature; they are designed. So the information still requires a non-foundational force (human ingenuity) to create the tiles which are used to create information in the pattern “I love you.”

Even then, the odds that “I Love You” would form are quite rare. Assuming an equal sample of each letter of the alphabet (as well as an infinite supply of them), you have 8 letters, so the odds of pulling these particular letters would be 1/26⁸, or 1 in 208,827,064,576, which is: 4.79 x 10 ^-12. If you include the space as a character, we have 10 characters and 27 possibilities each draw: 1/27¹⁰, or 4.9 x 10^-15.

In reality, however, Scrabble boxes do not contain an equal sampling of each letter. Instead you have 12 Es; 9 As & Is; 8 Os; 6 Ns, Rs, Ts; 4 Ds, Ls, Ss, Us; 3 Gs; 2 Bs, Cs, Fs, Hs, Ms, Ps, Vs, Ws, and Ys; 1 J, K, Q, X, Z. Finally, there are 2 blanks. This yields 100 total pieces. If we use the blanks as spaces, the odds for each letter in “I [blank] Love [blank] you” are:

I = 9/100
Blank = 2/99
L = 4/98
O = 8/97
V = 2/96
E = 12/95
Blank = 1/94
Y = 2/93
O = 7/92
U = 4/91

Because we are not dealing with an infinite number of tiles, we have to reduce how many are available after each selection. Thus, we have a 9/100 chance of pulling an I on the first draw from the box. If we do so, there are now only 99 tiles remaining, 2 of which will be blanks. That means we have a 2/99 shot for the blank, etc. Note that when a letter repeats (for instance, the O), we have to decrease the number remaining too. Thus, the first draw of an O is 8/97 but the second is 7/92 (because the first draw picks one of the Os). Finally, we get the combined odds by the following:

9/100 x 2/99 x 4/98 x 8/97 x 2/96 x 12/95 x 1/94 x 2/93 x 7/92 x 4/91, which is:

774,144 / 62,815,650,955,529,472,000

Or 1.23 x 10^-14

Which is roughly 1 in 81 trillion. So even though the tiles were created by humans, a random arrangement of them to spell out “I love you” is still extremely rare.

The above does, however, help us understand a bit about DNA. As most are already aware, DNA uses 3-base codons to create amino acids. There are four possible DNA bases (ACGT), and that means that means 4³ (64) possible combinations of those letters. However, there are only 20 amino acids. As a result, amino acids are often encoded by multiple numbers of codons. For instance, Leucine (L) can be encoded by CTT, CTC, CTA, CTG, TTA, and TTG. Which means there are 6 possibilities for L. In fact, quickly going through the amino acids (using their single-letter code name) we find:

I = 3
L = 6
V = 4
F = 2
M = 1
C = 2
A = 4
G = 4
P = 4
T = 4
S = 6
Y = 2
W = 1
Q = 2
N = 2
H = 2
E = 2
D = 2
K = 2
R = 6
Stop = 3

As you can see, all 64 possible combinations would be represented in the above. Therefore, we can say that given a random piece of DNA with 3 codons, there is a 3/64 chance that it is I (Isoleucine) and a 2/64 chance that it is N (Asparagine), etc.

Because base pairs are so prevalent, we can treat them as if there is an infinite supply of them. As a result, if we wanted to calculate what the odds would be that six base pairs will code for Isoleucine and then Asparagine, we would simply multiply 3/64 and 2/64 to yield: 6/4096, or about 1 in 683.

Of course, proteins can have hundreds of amino acids chained together in polypeptides. (In fact, by convention, most scientists do not consider a polypeptide chain to be a protein until it has at least 50 amino acids in it, although that is an arbitrary dividing line.) Because of their size, the odds of even a single 50-amino acid polypeptide forming are quite rare. In fact, even if they were simply a chain of L (Leucine), which has a 6/64 chance of forming for each L, the odds of 50 formulating would be 6⁵⁰ / 64⁵⁰, which is roughly 8 x 10³⁸ / 2 x 10⁹⁰ which is approximately 4 x 10^-52, or 1 chance in 3 x 10⁵¹.

Clearly, this method of explaining DNA is insufficient to explain even a basic protein, let alone complex cells and higher organisms.

This brings us to our next point, which is something that Mighty Pile brought up: the definition of information (i.e., something that is non-repeating, non-random, and not based on foundational forces) seems to exclude the ability of random, non-directed processes in the first place. As such, B) seems to be proven by stipulation, which means it relies on a circular argument.

However, when we examine B) carefully we see that it does not rely on circular reasoning when cashed out. To demonstrate how that is possible, I must first point out that the Darwinist must assert the opposite of B). They must assert that information can arise from random, non-directed processes (as evidenced by premise C)). And this is demonstrated by the fact that you are reading this blog post, which is information.

This blog post has an author. The author is not a random, non-directed process. But, if Darwinism is correct, at some point we can link my existence back to a random, non-directed process. Therefore, in a causative sense, the Darwinist would say that a random, non-directed process somehow created a non-random, directed process that was able to create information.

And it is because this option remains open to the Darwinist that B) does not entail circular reasoning. All the Darwinist needs to do is to show that Information can arise from forces that are non-random, non-repetitive (to exclude crystals) and non-foundational if those forces (we will call them meta-forces) are themselves built on random, non-directed forces. In other words, the Darwinist can argue: “Information comes from meta-forces, which are non-foundational; but meta-forces come from foundational forces.” Putting it into this two-step process would avoid the circular reasoning charge, while also giving the Darwinist a possible route to establishing C).

So the question now becomes, can random, non-directed processes create non-random and non-repeating meta-processes that could then create information in the form of DNA? DNA is one of the simplest information processes we can think of (compare it to trying to establish the framework for a spoken language), but even it is vastly complicated. In order for DNA to function, it has to store information that is used to create amino acids that bond together to form proteins that then create the mechanism for storing and reading DNA. In other words, in order for DNA to function biologically, we need to have a loop where DNA is used to create the processes needed to create more DNA. DNA is copied via cellular processes that are created with proteins that are themselves created by DNA. Thus, we have a vicious cycle going on.

But before we get to the loop, is there a simple way to just encode amino acids into DNA? Amino acids, after all, are fairly easy to create in a test tube, as Stanley Miller demonstrated (albeit his experiment does not prove what he thought it proved). Using those same “primitive” conditions, however, it is not possible to create DNA.

DNA also presents a problem because, as you’ve seen above, sometimes as many as six different DNA codons can represent a single amino acid. While moving from a DNA codon to an amino acid is easy, moving from the amino acid to a particular strand of DNA is much harder.

Due to the limitations of DNA, Francis Crick proposed that life began based on RNA instead of DNA. RNA is only single stranded, as opposed to the DNA double helix. RNA can also sometimes function similarly to proteins. DNA, however, is much more stable and less prone to errors (which is why an intelligent being would pick DNA instead of RNA to start life off; and which is why Darwinists claim DNA was “selected for” by Natural Selection).

Which brings up an important point. The “central dogma” (as Crick named it) is DNA to RNA to protein. It doesn’t go in the opposite order. (There are a few exceptions to the strictness of the “central dogma”, most notably RNA viruses (like HIV) which go from a single strand of RNA to DNA before then going through the “central dogma”; but there are no instances that I am aware of where proteins go to RNA then to DNA.) This makes it highly unlikely that amino acids bonded to become proteins and then those proteins created RNA that was then made into DNA and eventually stored in cells.

That means we had to start someway with DNA or RNA and then create proteins from that; but in order to create the proteins, it means we must have the structure in place by which RNA can be converted to a protein. Once again, we’re left with the chicken and the egg problem. And this system cannot have arisen by blind chance, since as you’ve seen even a single protein of 50 of the most common amino acids has astronomically long odds at forming randomly.

Regardless of where we start, we have to have some method of going from a random soup of amino acids to a particular sequence of amino acids being coded in information, be it RNA or DNA. But this will only start to happen if there is a reason for the information of a protein’s make-up to be converted to RNA or DNA.

That DNA is useful for life is not debated. Suppose that the amino acid “soup” manages to create a protein that could be used by a cell later on. It would be useful for the cell to have a way to rapidly create this protein. And the protein is created from amino acids that can be stored in DNA. Obviously, if we have this end in mind, we could design the process by which the DNA code comes about. But this requires teleology, which Darwinism denies. We cannot have the end of a working cell in mind; we have to have completely random processes that somehow create the necessary steps involved.

But suppose that we are left with only the random creation of the system to begin the evolutionary process. According to modern materialistic theory, life first became possible about 3.5 billion years ago. That is, the Earth cooled enough, the atmosphere was in the correct state, water existed, etc. so that life would not be extinguished if it was formed. Amazingly enough, according to these same scientists, the first life on Earth appeared roughly 3.5 billion years ago. In other words, as soon as it was possible for life to exist on Earth, life did exist on Earth. This must mean that the creation of life ought to be an “easy” process, given materialistic claims. If it is easy, then it should not rely on a process that has such poor odds of succeeding. Either life’s occurrence on Earth was a miracle against all odds, or else this cannot be how life began on Earth.

NOTE: This post has been updated since it was originally posted to correct the line: “While this would be a random process creating information, it is not using foundational forces” to “While this would be a semi-random process creating information, it is not using foundational forces.”

I mentioned in my comments with Mighty Pile the Gambler’s Ruin. The GR occurs when a gambler runs completely out of money. There are two aspects of the GR that impact our understanding of Natural Selection. First is the fact that if you are at a numerical disadvantage, then even if you have a statistical advantage in gambling you will often hit GR first simply because the other person can take “more damage” before he reaches it. Thus, just because one individual gains a favorable mutation does not mean that that mutation will be automatically chosen for due to the sheer number of competitors that the individual would have to compete with.

But more importantly is the fact that Natural Selection, in order to work at all, is an All-Or-Nothing proposition. That is, favorable traits must be selected for while unfavorable traits must die out. In one of his comments, Mighty Pile said:

Some traits DO confer an advantage to a particular organism and its progeny. While fit individuals certainly do die sometimes and unfit individuals certainly do live sometimes, the fit organisms would outcompete the unfit ones in large numbers. One antelope’s chance vs another antelope’s chance may be a 49%-51% split. But in a whole herd, the one that gets eaten will almost always be of the slower variety, or of the sick or injured variety. This is, of course, supposition based on logic. I don’t know how I’d prove it right now; it seems obvious in any context I can come up with for it. The difference between fit and unfit would probably be very small most times, setting up a sort of tipping point situation. I don’t have to outrun the bear, I only have to outrun you, right?

In response, I pointed out that one would be foolish to wager everything he owned for a chance to win a billion dollars if he only had a 51% chance of winning it and a 49% chance of losing it. This did get me to thinking a bit further, however, and I developed the following.

Suppose you start with 100 individuals. Each begins with $100. Each wagers $100 in order to gain $100. The odds are 51% win and 49% lose each bet, but with the following stipulation: as soon as you hit $0, you’re out of the game. You cannot continue. This is important to mimic Natural Selection, because as soon as you die you can no longer reproduce. It’s over. So you need a final set point.

How long will it take for a person to reach $1,000 given this structure? And how many people will hit GR before that occurs?

I made up an Excel spreadsheet to show this to me (click here for graphic). It assumes a literal 51% - 49% split for each round (in other words, I don’t randomize the data; this is the “ideal”). The vertical axis is how much money people have; the horizontal axis is the number of rounds. The number plotted in each cell is how many people remain for each row (i.e., how many people have whatever money is in that row). The bottom line beneath the graph simply sums how many people remain (i.e., those who did not hit GR). The cell at the far right of the 0 line is the grand total of those who hit GR.

Thus, we begin with 100 people holding $100. After the first round, 49 people are bankrupt and 51 people have $200. In the second round, 51% of those 51 people (26.01) at $200 will gain another $100 for a total of $300, while 49% (24.99) lose $100 to go back to $100 total. For the third round, we again calculate each group: 51% of the 26.01 (13.27) at $300 will go to $400; simultaneously, 49% of the 26.01 (12.74) drop back to $200. In the meantime, 51% of the 24.99 (12.74) who dropped to $100 will gain $100 and make it back to $200. They combine with the 12.74 who lost $100 to drop down to $200 to make 25.49. Finally, 49% of the 24.99 (12.25) who were at $100 will go bust.

Note that unlike in real life (where a whole number of people either win or lose), these calculations are made with the decimal points from the previous numbers still intact. In fact, I used dependent formulas for each cell. If we were rounding before we did the math, the answers would vary slightly.

And the results: It takes 11 rounds for the first person to hit $1,000 (and that’s only if you round 0.58 up to 1; the line does reach in round 9, but the value would round down to 0). In the meantime, 75.93 people have gone bust. That means that in order to get one person from $100 to $1,000, 76 people have to go bankrupt. And that’s starting with 100 people. A 1% advantage does not provide much of an advantage at all under these circumstances.

Natural Selection falls to the same principal. Just because a favorable mutation may confer a 1% advantage onto an antelope does not mean that the antelope really has that much more of an advantage than other antelope. And I should point out that living systems are actually far more complex than even this illustrates.

The key to why this works this way is because the chart is capped at 0. Once you hit 0, it’s over. That provides a literal line in the sand that has a huge impact. Because in Natural Selection death is such a line in the sand, this demonstrates that even a 1% advantage holds no real benefit to the furtherance of a trait in the species.

In reality, survival rarely comes down to a single trait though. Chance encounters are almost always going to outweigh any mutational advantage of a single trait. Consider all the following that mitigate against the classical view of Natural Selection:

* An antelope is born with 1% more speed than any other antelopes who have been born. However, when the antelope is a newborn, he is not as fast as the adults. As a result, despite being 1% faster than all other newborns, he is still slower than the slowest adult; therefore, he remains a preferential target for predators. If he is near adults at the edge of the herd when lions attack, they will go after him rather than the adults. This brings to mind the second point:

* As Mighty Pile pointed out, there is an oft repeated joke that one need only be faster than the slowest prey when a predator attacks. This, however, ignores the fact that if you are faster than me, but you are five feet away from a hungry bear while I am a quarter mile away from the hungry bear, the bear will catch you before you can run far enough to surpass me and make me a target.

* Sometimes pure dumb luck happens. A ram may be the fittest ram ever, but if he slips and breaks his leg, he’ll be eaten. And accidents happen quite often in nature. And even aside from nature. A highly specialized and advanced snake in Baghdad might happen to get hit by a mortar round fired from an insurgent that was not intended to strike the snake, but did. Or a random lightning strike could kill an elk in the forest who was “superior” to the other elk. When it comes to random events, traits have no bearing on survivability. There is no survivability trait for bad luck.

* For that matter, the strongest bull may be cut down by a viral infection that attacks only strong animals, leaving the weak bulls alive. The weak bulls are “more fit” (by definition, since they survived) but once the infection runs its course the herd would have been better off with the stronger bulls.

* A mutation for greater intelligence might occur in a sheep that’s also the least hearty sheep in the herd. Despite the fact that this intelligence trait would benefit the herd as a whole, the sheep dies of an illness before reproducing.

So survival rarely is about any one trait anyway. Instead, to have the best chance at surviving, organisms need to have a wide range of traits, any one of which may or may not be relevant at any particular time. But some traits are mutually exclusive. Because evolution must be blind (in a materialistic world) it cannot predict which trait will be needed in the future. And because it cannot predict what is needed (after all, it is non-teleological; and furthermore, even intelligent agents like weathermen cannot predict what will happen in the environment tomorrow), the random forces of nature will far outweigh any slight statistical advantage that individuals in a herd have.

So the only way to have beneficial mutations that avoid the GR problem is if they grant a far greater than 1% chance upon the individuals (after all, think of mutations, which convey far more than a 1% disadvantage to the individuals and therefore are seen!), or if they occur more often than random mutations would enable them to occur so that more individuals get the trait (remember, we started the above graph with 100 individuals already having $100, and 76 of them went bankrupt before a single person reached $1,000; if you had 1,000 people to begin with, 760 would go bankrupt…but you’d have 10 make it to the $1,000 mark, so clearly having more individuals get the same mutation would help), or the mutation would have to occur in an individual that is already “more fit” due to other traits to begin with (and that brings up the converse: a detrimental mutation can occur in those who are “more fit” due to other traits and therefore be “selected for” simply because it’s riding along with the system; whereas a “less fit” organism might evolve a wonderful trait that cannot overcome the aspects that make it “less fit” and therefore that trait is not “selected for”).

That’s a lot of front-loading you need before you can get the system going. Living systems are far too complex to be affected greatly by any slight advantage in a single trait.

(Click on graphic to have it open in a new browser window if it doesn’t display fully for you.)

Okay, after having worked on the equations for a bit more (and discovering I had left off an important set of parentheses in the previous function formulas), I have found the simplified version of each of them. These equations will once again use two variables, but since the 6n +/- 1 format is already established for primes, I’ve reworked it. For the following, n = the row you’re trying to build on the graph. This is slightly misleading because each row is actually patterned off of the 6n +/- 1 format itself. Therefore, there are two rows for each n. The 6n - 1 and the 6n + 1 value. And for purposes of the chart, a 6n - 1 number is black and a 6n + 1 number is red.

Finally, there is a controlling x value that determines how far to the right you’ll place the cell. Basically, if you were testing for a prime, you could loop x starting at 0 and running the equation, then increasing x by 1 and running the equation, repeating until x equals the n value you’re searching for. If you’re testing a 6n - 1 number, then any black values in the n column will be factors; if you’re testing a 6n + 1 number, then any red values in the n column will be factors.

Now are you ready for the massively complex new equations?

x(6n - 1) + n ; (Use to find black values on the 6n - 1 line.)
x(6n + 1) + n ; (Use to find red values on the 6n + 1 line.)
x(6n + 1) - n ; (Use to find red values on the 6n - 1 line. x must be greater than 0.)
x(6n - 1) - n ; (Use to find black values on the 6n + 1 line. x must be greater than 0.)

And just to demonstrate it, here’s the values for the first couple of rows.
n = 1
x = 0

Black finds black @ 1
Red finds red @ 1
Red finds black @ - 1*
Black find red @ - 1*

n = 1
x = 1

Black finds black @ 6
Red find red @ 8
Red find black @ 6
Black finds red @ 4

n = 2
x = 0

Black finds black @ 2
Red finds red @ 2
Red finds black @ - 2*
Black finds red @ - 2*

n = 2
x = 1

Black finds black @ 13
Red finds red @ 15
Red finds black @ 11
Black finds red @ 9

* = why X must be greater than 0 for the final two equations.

By the way, the first two equations are equivalent to the equations that I came up with before if n = S + 1. The second two would not be due to some misplaced parentheses in my original formulas :-( Oh well. This way is simpler and more “elegant.”

Note: X must be > 0 to work correctly.

F(S, x) = S + (1 + x) + x[(4S + 2) + 1 + (2S + 1)].
F(S, x) = S + x + 1 + x[(3(2s + 1) + 1]
F(S, x) = S + x + 1 + x[(6S + 3) + 1]
F(S, x) = S + x + 1 + (6sx + 3x + x]
F(S, x) = S + 2x + 1 + 6Sx + 3x
F(S, x) = S + 5x + 6Sx + 1

F(S, x) = S + 1 + x(4s + 4) + 1 + (x – 1)(2S + 1)
F(S, x) = S + 1 + 4sx + 4x + 1 + 2sx + x - 2S - 1
F(S, x) = S + 1 + 6Sx + 5x - 2S
F(S, x) = 6Sx + 5x + 1 - S

F(S, x) = S + (1 + x) + x[(4S + 4) + 1 + (2S + 1)]
F(S, x) = S + 1 + x + x(4S + 4) + x + x(2S + 1)
F(S, x) = S + 1 + x + 4Sx + 4x + x + 2Sx + x
F(S, x) = S + 1 + 7x + 6Sx

F(S, x) = S + 1 + x(4S + 2) + 1 + (x-1)(2S + 1)
F(S, x) = S + 1 + 4Sx + 2x + 1 + 2Sx + x - 2S - 1
F(S, x) = -S + 6Sx + 3x + 1
F(S, x) = 6Sx + 3x + 1 - S

(6n - 1) functions:
F(S, x) = S + 5x + 6Sx + 1
F(S, x) = 6Sx + 5x + 1 - S

(6n + 1) functions:
F(S, x) = S + 1 + 7x + 6Sx
F(S, x) = 6Sx + 3x + 1 - S

Rewritten:

(6n - 1) function:
F(S, x) = 6Sx + 5x + S + 1
F(S, x) = 6Sx + 5x - S + 1

(6n + 1) function:
F(S, x) = 6Sx + 7x + S + 1
F(S, x) = 6sx + 3x - S + 1

What would happen if you made a chart based on the following patterns (wherein everything repeats except for the “Initial Skip” which is a one-time event):

Etc.

You would get a graphic like this one:

You might be thinking, “So?” Well, one of the things that I enjoy about mathematics is that it is easy to come up with several different paths to the same destination. In fact, I remember my geometry teacher in high school say the following about someone’s proof: “Your method gets us to the solution, but it’s like driving from here [a small town near Pueblo , Colorado ] to Colorado Springs via Fairbanks , Alaska . You get there, but it’s the long way.”

Mathematicians typically seek the most “elegant” solution, where elegance is defined as the simplest solution. You can have bulky proofs that get you where you need to go, but a sleek, elegant proof is preferable. However, in my opinion while they may be preferable in the ethereal sense, the various perspectives on mathematical problems actually help us to more fully understand the concepts involved. And indeed some mathematical truths may be easier to see in one perspective than in another perspective.
That is one of the reasons that I’ve been looking at the “Factor Field” sheet I’ve made in Excel and drawing conclusions from it. Now I’ll be the first to admit that I’m never very good at calculations in mathematics, but I am a very visual mathematician (this is why I’ve always liked geometry the most out of all kinds of math). Some who are really good at calculations have already proven such things as: All prime numbers exist in the format of 6n +/- 1. They proved this by showing that any other number than 6n +/- 1 must be divisible by 2 and or 3, and therefore cannot be prime. I came to this same conclusion independently by looking at the Factor Field and seeing the 6-spike there and wondering what it would look like if I converted it to base 6 numbers.

I think the visual representations are therefore a very powerful tool. Again, I’m not the greatest at calculations and therefore if I had to limit myself to calculations I wouldn’t get very far. But because in our day and age computers make it easy to come up with visual representations, someone like me can look at these patterns and then formulate equations just as complicated as those who are good at calculations come up with. That’s what I did with the above graphic. So how did I come up with it?
Well, remember this graphic?

I started by looking at the 5 column. Then I counted the 6-spikes and put on a graph whether the number was higher or lower than the 6-spike. So for the first one, the 5 line starts with the number higher. It got a black cell. After that, the 5 column skips 2 6-spikes before touching lower on the 6-spike. Therefore, I skipped 2 and then put a red cell. After that, it skips 1 6-spike (where it’s actually on the 6-spike) and repeats the sequence.

I did the same with 7, 11, 13, etc. until I got the above chart. Here’s it reproduced with the vertical and horizontal axis labeled appropriately as well as a “total” line at the bottom:

Now here’s how to understand what is there. Look at the column headers. They are all multiples of 6. Those who remember that P numbers (as I’ve defined them) are in the format of 6n +/- 1 should already know where this one is heading. A cell will be black if it is in the 6n – 1 format; red if it’s in the 6n + 1 format.

The total line is created in this way. Excluding the first red and black (the cells that appear after the initial skip, but not after any of the repetitions), if there is a red cell (or more than one) and no black cells in the column, the cell for that column on the total line gets marked red. If there is a black cell (or more than one) and no red cells, it gets marked black. If there are both red and black cells, it gets marked blue. If there are no red or black cells, it remains gray.

Now if there is a red cell on the total line, that means that the 6n + 1 number is not prime. If it’s black, then 6n – 1 is not prime. If it’s blue, then neither 6n + 1 nor 6n – 1 are prime. If it’s gray, then both 6n + 1 and 6n – 1 are prime.

Pick one of the numbers on the total line (let’s use 84). The cell on 84 is red. Red refers to a 6n + 1 number. Therefore, we know that 84 + 1 is not prime. And sure enough, 85 is not prime. Furthermore, you can look in the column above the total line and see the factors of 85 listed out in all the red cells: 17 & 5.

Since 84 is red (not blue) then we know that the 6n – 1 of 84 (83) is prime.

Let’s look at another number: 120. 120 is blue, so we know that neither 119 nor 121 is prime. And sure enough, we can find the factors of each in the above graph. 119 has factors of 17 and 7 (119 is a 6n – 1 number, so look for the corresponding black cells) and 121 has 11 as a factor (121 is a 6n + 1 number, so look for the corresponding red cell).

Now here’s the thing about this graph. I made it without doing a single factor calculation! Instead, I looked at the pattern, and the pattern shows us this (where a “-“ refers to a 6n – 1 number and a “+” refers to a 6n + 1 number, represented by black and red respectively):

5:   0 - 2  + 1
7:   0 + 4  - 1
11:  1 - 6  + 3
13:  1 + 8  - 3
17:  2 - 10 + 5
19:  2 + 12 - 5
23:  3 - 14 + 7
25:  3 + 16 - 7

And furthermore, from looking at this pattern, I’ve been able to deduce a general rule whereby you can figure out what the repetition of any specific line (L) on the graph would look like. To find it, you do the following:

1. Calculate if number N is a 5 or 1 number (i.e. if it is in the 6n + 1 or 6n – 1 format).
A. Convert to base-6, see if last digit is 5 or 1 or…
B. Find the remainder of N/6 (you can use N mod 6 on a calculator). It must be either 5 or 1.

E.g. 35. 35/6 = 5 r 5. It is a 5 number.
E.g. 49. 49/6 = 8 r 1. It is a 1 number.

2. Calculate Initial skip for starting skip variable (S).
A. If 5 number, the integer (non remainder) portion of N/6 is the Initial Skip: S = N/6
B. If 1 number, the integer (non remainder) portion of N/6 less 1 is the Initial Skip. S = (N/6) - 1

E.g. 35. 35/6 = 5 r 5. 5 is Initial Skip.
E.g. 49. 49/6 = 8 r 1. 8 - 1 = 7. 7 is Initial Skip.

3. If N is a 5 number (Black), the pattern is: S + Black + (4S + 2) + Red + (2S + 1).

E.g. N = 35. S is the integer portion of 35/6 = 5. Substitute into the equation and you get 5 + Black + 22 + Red + 11.

4. If N is a 1 number (Red), the pattern is: S + Red + (4S + 4) + Black + (2S + 1).

E.G. N = 37. S is the integer portion of 37/6 less 1 (i.e. 6 – 1 = 5). Substitute into the equation and you get: 5 + Red + 24 + Black + 11.

The two patterns are:

S + Black + (4S + 2) + Red + (2S + 1)
S + Red + (4S + 4) + Black + (2S + 1)

In reality, the initial S is not part of the pattern, it just slides the pattern over. If we put the portion that repeats into brackets [], you get:

S + [Black + (4S + 2) + Red + (2S + 1)]
S + [Red + (4S + 4) + Black + (2S + 1)]

Repeat what’s in the brackets for however many times you need.

Now as you can see from the graph, you don’t really need to check many of the values to determine if a specific number is prime or not. You only need to find the column that the number appears in after sliding over from the left and make sure that there are no corresponding marks in the column above it.

So let’s start with one that we can empirically verify on the chart. The number 35. 35 is a “Black” number because it is a 5 number (as shown above). The Initial Skip value (S) is 5. This means that there are 5 skips before the number appears, so the number will appear in the 6th column. So you can always find the column (C) by finding S + 1. In our test, C = 6.

Now since 35 is a “Black” number (i.e., a 6n – 1 class number) we only have to worry if there are any other “Black” numbers in the same column. Red numbers do not matter because they will not match up with the Black numbers. However, there will be two functions that we have to use in order to test a specific value. The first function is the one that calculates where on the grid a Black number places a Black mark. The second function calculates where on the grid a Red number places a Black mark (after all, they alternate back and forth as they go across).

So when testing where a Black number puts a Black mark on the grid, we use this function:

F(S, x) = S + (1 + x) + x[(4S + 2) + 1 + (2S + 1)].

To us this function, we’ll need a nested loop (a loop within a loop). First start with S and x both equaling 0. We can set up the test in the following computer pseudo-code:

For S = 0 to C
X = 1
            While TestValue < C
                        TestValue = F(S, x)
                        x = x + 1
            End While Loop
If TestValue = C, then the number is not prime and you should break loop.  Otherwise…
Next S

By the way, if you’re wondering why we start with x = 1 instead of x = 0, it’s because whenever x = 0 then the function value will always equal the S value + 1. And therefore, F(S, x =0) can be ignored because it will always be less than the C value except when it equals the C value (which means you’re testing the number against itself, which is pointless).

Let’s go ahead and test it with our variables as x = 0 and S = 0 to show this. Remember that when S = 0 that means the initial skip value is 0. Since we’re looking at Black numbers, we know that this corresponds to the first row on the chart, so we can check that to verify if the function works correctly. We are therefore going to re-create the first row, but we only need to do so up until we get to column 6 since the number N that we are testing is 35.

F(S, x) = S + (1 + x) + x[(4S + 2) + 1 + (2S + 1)].
F(0,0) = 0 + (1 + 0) + 0{[4(0) + 2] + 1 + [2(0) + 1])}
F(0,0) = 1

When x = 0, therefore, we see that F(S, x = 0) will equal S + 1. And this is accurate: The first Black mark does indeed show up in the first column. (You can multiply the F(S,x) answer by 6 to get the correct header if you want, but the labels aren’t actually relevant here.)

Anyway, let’s continue with x = 1:

F(S, x) = S + (1 + x) + x[(4S + 2) + 1 + (2S + 1)].
F(0,1) = 0 + (1 + 1) + 1{[4(0) + 2] + 1 +[2(0) + 1]}
F(0,1) = 2 + (2 + 1 + 1)
F(0,1) = 6

At this point F(0,1) is not less than 6 because it in fact equals 6. Therefore, the loop breaks out. The program then tests if the final value equals the value being searched for. It does. Therefore, 35 is not prime and the program terminates.

Just to show it works, let’s see what happens if we pick a prime “Black” number, such as 29. First, we calculate the S number. S = the integer portion of 29/6 = 4.

C = S + 1, so C = 5.

Now we start through the function, once against starting with S =0 and x = 1. Obviously, the first result of F(0,1) will yield the exact same results as what is shown above. Since the C value is now 5, however, we have F(0,1) = 6 which is greater than 5 but it is not equal to 5. This means we increase the S value to 1 and reset the x to 1 and repeat the loop:

F(S, x) = S + (1 + x) + x[(4S + 2) + 1 + (2S + 1)].
F(1,1) = 1 + (1 + 1) + 1[(4(1) + 2) + 1 + (2(1) + 1)]
F(1,1) = 3 + [6 + 1 + 3]
F(1,1) = 13.

If you check the chart, you will see that the second black mark for the 11 row is indeed located at column 13.

Now we can see that {F(1,1) = 13} greater than {C =5}. Therefore, we increase the S to 2, and reset the x to 1.

F(S, x) = S + (1 + x) + x[(4S + 2) + 1 + (2S + 1)].
F(2,1) = 2 + (1 + 1) + 1[(4(2) + 2) + 1 + (2(2) + 1)]
F(2,1) = 4 + (10 + 1 + 5)
F(2,1) = 20.

{F(2,1) = 20} greater than {C = 5} so increase S to 3 and reset x to 1.

F(S, x) = S + (1 + x) + x[(4S + 2) + 1 + (2S + 1)].
F(3,1) = 3 + (1 + 1) + 1[(4(3) + 2) + 1 + (2(3) + 1)].
F(3,1) = 5 + (14 + 1 + 7)]
F(3,1) = 27.

Now I’m sure you can see the pattern that is resulting from this function. Increasing the S value will always yield a larger answer (in fact it’s the previous answer + 7). Therefore, if at any point when x = 1 the function is greater than the C value, we know that there are no more factors that we need to test. Note that this wouldn’t prove that the number is prime; it merely proves that no other Black numbers are factors for it. We still have to do a function for where the Black marks appear on the Red numbers, and we get that in the following function:

F(S, x) = S + 1 + x(4s + 4) + 1 + (x – 1)(2S + 1)

Once again, we can test it with figures we know from the chart. For instance, when S = 0 and x = 1:

F(0,1) = 0 + 1 + 1(0 + 4) + 1 + (0)(0 + 1)
F(0,1) = 6

If you look at the chart, the first “Red” number is on the 7 line, and the first black mark is indeed at column 6. Testing S = 1 and x = 1, we see:

F(1,1) = 1 + 1 + 1(4(1) + 4) + 1 + (0)(2(1) + 1)
F(1,1) = 1 + 1 + 8 + 1
F(1,1) = 11.

The first black mark on the second Red number (13) is indeed at column 11.

So that means that we can test any “Black” number (i.e. 6n – 1) to see if it’s prime by running it through these two functions:

F(S, x) = S + (1 + x) + x[(4S + 2) + 1 + (2S + 1)]
F(S, x) = S + 1 + x(4s + 4) + 1 + (x – 1)(2S + 1)

Testing for “Red” numbers (i.e. 6n + 1) is similar. In those cases, the functions look like this:

F(S, x) = S + (1 + x) + x[(4S + 4) + 1 + (2S + 1)]
F(S, x) = S + 1 + x(4S + 2) + 1 + (x-1)(2S + 1)

You can test them to see they match up to the chart if you want. I will simply conclude this by pointing out that using math no more complex than simple Algebra, you can test whether a number is prime or not…all without doing massive amounts of division! The only division you have to do is to divide your test number by six to get the starting integer (so you can get the S value) and the remainder (so you can determine if you’re looking for a 6n + 1 or 6n – 1 number).

Even if you only casually read through news websites (such as those of CNN or FOXNews), several times per month you will notice headlines such as the following:

TOO MUCH, TOO LITTLE SLEEP TIED TO ILL HEALTH IN CDC STUDY

Study: Long-Term Breast-Feeding Will Raise Child’s IQ

WOMEN, WANT A HEALTHY MARRIAGE? MARRY MAN UGLIER THAN YOU, STUDY SAYS

STUDY: FOOD IN MCDONALD’S WRAPPER TASTES BETTER TO KIDS

Study: 1 in 50 U.S. babies abused, neglected in 2006

And naturally we’re all aware of the competing studies that exist too. One study shows that eggs are bad for you; another that they’re good for you. One study shows how margarine is a healthier alternative than butter; another that butter is better for you. With so many competing studies, you can find a scientific backing for just about any position you want to take (especially in health matters).

The existence of so many studies helps to emphasize a point regarding statistical analysis. Despite being a powerful tool, if you do not set up the guidelines and restrictions for your samples properly any statistics you observe won’t amount to a hill of beans. And we’re not even talking about the inherent fluctuations that require the existence of error bars (that’s the line that says +/- 3%, for example). Nor are we even addressing political manipulation of statistics in the form of pollaganda. Instead, I’m talking about something at the heart of statistics itself—it’s a universal.

To demonstrate what it is, let us first ask a simple question. When we do a statistical analysis of some observation, for what reason are we doing it? As you can see in the above headline examples, most of the time studies are done to find a causal linking between some object and/or action and some result. Thus, the first headline above says that too much or too little sleep (the cause) is “tied” to “ill health” (the effect). We also see that women should marry uglier men for a healthy marriage (in a study obviously written by an ugly man).

Now let us assume that there is a correlation that all these studies found. Let us assume that it is the case that people who sleep less than six hours a night weigh more than those who sleep eight hours a night, and that women who married uglier men (however that is defined) are in healthier (however that is defined) marriages. The fact of the matter is that when you compare any subset of a group, however you wish to define that subset, with the rest of the group as a whole, you will find things that the small group has in common at a statistically higher rate than the group as a whole. This happens automatically and does not mean that it is relevant in a causative sense!

To give a simple example, let’s examine hockey (since I like hockey). There are 30 teams in the NHL. Of those 30 teams, 7 are named after animals (the Penguins, Bruins, Thrashers, Panthers, Ducks, Coyotes, and Sharks) and 7 are named after people-groups (the Islanders, Rangers, Canadiens, Senators, Blackhawks, Oilers, and Kings). Each group of 7 constitutes 23% of the teams in the League.

There have been 80 Stanley Cups awarded since 1926. During that time, teams named after animals have won 8 Stanley Cups, which means that they won 10%. However, teams named after people-groups have won 39 Stanley Cups during that time, which means they won 49% of them. Clearly, having a team named after a people-group instead of after an animal provides a statistical advantage to a hockey team…

Perhaps someone could argue that the statistical data isn’t fair. After all, the Thrashers (1999), Panthers (1993), Ducks (1993), Coyotes (1996), and Sharks (1991) are all teams that did not exist before the 1990s! On the other hand, the Rangers, Canadiens, Senators, and Blackhawks all existed in 1926 (the start of this survey). Furthermore, the Kings were founded in 1967, the Oilers in 1971 and the Islanders in 1972. Of the animal teams, only the Bruins were around in 1926 (the Penguins were founded in 1967). Thus, using 1926 as the baseline (since before that there were other teams besides just NHL teams that could play for the Cup), the average year of founding for animal teams is 1981 and for people-group teams it’s 1945.

However, we can adjust for that. Animal teams have won a Cup on average every 3.25 years they’ve existed; while people-groups win a Cup for every 1.59 years they’ve existed. Clearly, it still remains better to have a team named after a people-group than an animal. (And I’m not biased since I cheer for the Avalanche, which is neither a people-group nor an animal…)

Now here’s the thing. The statistical data that I’ve given here is all correct (assuming I didn’t make any typos or anything of that nature), but every rational person would immediately recognize that the type of name a sports team has, has no bearing on the performance of that team. This is an attribute that is linked statistically, but the statistical linkage is accidental rather than causative.

Every time that we do these surveys and examine the numbers we have to realize that there are some number of things that will be discovered in common that are accidental correlations. The problem is that we ignore most of these connections. And when I say we ignore them, I don’t mean that we test the data and then go, “This isn’t relevant” but we do not even look for them in the first place. After all, were it not for the fact that I was looking for an example for this blog entry I would never have cared what percentage of teams named after animals won the Stanley Cup. This correlation would have been excluded a priori as being irrelevant.

But these irrelevant correlations are important to statistical analysis! Why? Because since a certain percentage of linkages are accidental, we have to account for them in our conclusion. In other words, we have to have some way of determining if the link we discover is causative or if it is merely the kind of statistical fluke you get when examining hockey mascots. And that means that we would need to examine all possible connections and discard those that are accidental in order to find out if the statistical percentages are covered.

That, however, is impractical to the point of impossibility. After all, it is relatively easy to come up with statistical correlations between things. For instance, with my hockey example it took me all of 15 minutes to come up with that correlation. The longest part was pulling up the Wiki sheets on the number of Stanley Cup wins various teams had had. Indeed, based on my experience I would argue that it is so easy to come up with meaningless links between data that it will always remain more likely that a correlation is accidental than causative. That is, for every one true causative link between a subset of a group and the average of the entire group, I would argue there are several accidental links. And these accidental links are not always as obviously accidental as the examples I’ve given. (For a less obvious example, think of the correlation between diabetes and obesity. Does one cause the other? Or is it just a statistical fluke, similar to the names of hockey teams?)

If it is so difficult to prove our position statistically due to the possibility of accidental links, then what good is it to come up with a statistical correlation in the first place? For most studies that you read about in the media, the answer is: “None.” However, for scientists there remains one thing that a truly causative link can do that an accidental link cannot do that saves the field. A truly causative link will enable you to make a prediction that you can test and verify. If something is causative then it will continue to cause the effect at the same rate. On the other hand, if it is accidental then it is a random linkage, and random linkages will break down through further testing. For instance, the fact that people-group teams have won more Stanley Cups than animal teams does not help us predict who will win the Stanley Cup this year or next year or the year after that; therefore, it is an accidental link rather than a causative link. However, if further testing shows that the percentages of obese people who get diabetes remains constant, then we can have more confidence that that is a truly causative link rather than simply a statistical accident.

So there are some ways to salvage statistics. But it requires that we be able to conduct further tests with our predictions in place in order to sort out whether we have a meaningful causative link or a meaningless accidental link. If we cannot conduct those further tests, then any causative links will be lost in the noise of the countless accidental links. They may be true, but it is impossible to verify it.

By the way, here’s something I just thought of with the factor field. First, check out this graphic:

As you can see, the red cells are there to highlight gaps in the line with the lines that come off in both directions on the arms of the “starburst.” Now here’s the thing about that: wherever a factor exists on the 6-spike, it blocks the same area at whatever distance that is above and below the line. This means that if there’s a factor at 5, then 5 above and 5 below that 6-spike cannot be prime.

Here’s what I thought of. Whenever you have factorial numbers, then these arms will block primes from appearing. So if you have 10 factorial (10!), for instance (which would occur at 3628800 (which is 10 x 9 x 8 x 7 x 6…), then there cannot be any prime numbers occuring for 10 ahead and 10 after. It’s impossible, because the arms that branch off would block them.

So then…if you have 100!, then it blocks 100 before and 100 after. 1000! would block 1000 before and 1000 after. And so on. The bigger the number you’re doing the factorial for, the more numbers before and after get blocked. Which means there are huge sections of numbers that cannot possibly be prime.

Which really makes you wonder about infinity factorial… Because an infinite number of numbers before and after it would be blocked from being primes….

Mike Jones commented on my post over on the T-blog:

It’s actually not too difficult to show that all primes will end in 1 or 5 in base 6 (although not all numbers ending in 1 or 5 are primes), with the exception of 2 and 3.

We can categorically eliminate numbers that end in 0, 2, and 4. These are non-odd numbers. No prime other than 2 will be even.

That leaves 1, 3, and 5.

Let Xn be any number that ends with the digit 3 in base 6, except 3. This number may be represented as:

Xn = 3 + 6n, where n > 0

We can factor 3 out of the expression:

Xn = 3(1 + 6n), where n > 0

Therefore, all numbers ending in the digit 3 in base 6 are divisible by 3 => No prime numbers will end in 3 in base 6.

That leaves 1 and 5.

Your hypothesis is fact.

:D

This got me to thinking about my other hypothesis that all P numbers are either prime or have factors that are other P numbers. And now I’ve been able to prove that.

Remember that I previously defined a P number as: “The P class is defined as any number that is N +/- 1″ and N was defined as “a positive number that is divisible evenly by 6.” In reality, this definition is too restrictive, so allow me to redefine it slightly:

A P class number is any number in base-6 whose last digit (read left to right) is either a 1 or a 5.

Because this is a base-6 definition, everything in my previous definition still works (N is a multiple of 6 in base-10, so all N numbers will convert to ending in a 0 in base-6; N + 1 or N - 1 in base-10 will convert to a number that ends in either 1 or 5 in base-6 respectively).

Under this new definition of a P class number, we can also include the number 1 (as it is a number whose last digit is either a 1 or a 5).

Now, my hypothesis is that a P class number must be either prime or its factors must also be P class numbers. Since prime numbers are divisible by 1 and 1 is a P class number, we can actually exclude the “prime” portion from the above and simply say: A P class number can only be divisible by another P class number.

To prove this, we need to demonstrate that A) any P class number multiplied by another P class number will give us yet a third P class number (which I demonstrated in my previous post) and B) no non P class number can ever be multiplied by any other number (P class or not) to form a P class number. So to prove it:

1. Any number that ends in digit x multiplied by a number that ends in 1 will have x as it’s terminating digit. Therefore, if multiplying by a P class number that ends in 1 to get another P class number, you must multiply by a number that ends in either 1 or 5. Therefore, multiplying a P class number that ends in 1 requires multiplying by another P class number to yield a P class number.

2. In an even base system (and base 6 is an even base system since 6 is even), any number x multiplied by any even number will be an even number. Examples (in base-6):

3 x 2 = 10 (x = 3)
4 x 2 = 12 (x = 4)
12 x 4 = 52 (x = 12)

Therefore, any number multiplied by 0, 2, or 4 must be an even number and cannot end in either 1 or 5.

3. An even number multiplied by a number that ends in 3 will end in 0 in base-6 (we don’t really need to demonstrate this step, since even numbers were excluded in step 2, but I’m including it just to be complete); an odd number multiplied by a number that ends in 3 will end in 3 in base-6. Example:

2 x 3 = 10.
3 x 3 = 13.
4 x 3 = 20.
5 x 3 = 23.

Therefore, no number multiplied by a number ending in 3 could end in 1 or 5.

4. This leaves only 5. Taking a look at the multiplication table of 5 in base-6 is quite interesting:

5 x 1 = 5
5 x 2 = 14
5 x 3 = 23
5 x 4 = 32
5 x 5 = 41
5 x 10 = 50

If you note, this has the same type of thing we find in base-10 systems with the number 9. In base 10, you can construct the multiples of 9 by simply writing a column of numbers from 0 - 9. Then, to the right of each number, write the inverse (9 - 0) next to the previous number. Hence:

09
18
27
36
45
54
63
72
81
90

Here, the same thing is seen in base-6, only with the range of 0-5:

05
14
23
32
41
50

Likewise, just as the digits in the 9 sequence add up to 9 (i.e. 18 is 1 + 8 = 9, 36 is 3 + 6 = 9), so in base-6 in the 5 sequence the digits add up to 5 (23 is 2 + 3 = 5; 41 is 4 + 1 = 5). I’m guessing (though I haven’t tested it) that this is the case of any even base system. (I.e., given base N where N is an even number, then the multiplication table of N - 1 will display the above behavior.)

In any case, that’s just something I discovered but it’s not relevant to the current discussion. What is relevent is this: a number that ends in 5 can only end in 1 if multiplied by a number that ends in 5; and a number that ends in 5 can only end in 5 if multiplied by another number that ends in 1. Therefore, the only way to make another P class number multiplying by a P class that ends in 5 is if you multiply it by another P class number.

Conclusion: A P class number can only be divisible by another P class number.

I’m sure you’ve probably heard the phrase, “It just clicked into place.” Last night (or rather, very early this morning) I experienced that. Literally. Like it was an actual audible “click” sound as a realized something regarding the prime numbers in the “factor field” that I’ve developed in Excel.

To give some background, I’ve been conversing with someone via e-mail after my post the other day that included my reference to the factor field. This person has looked over my spreadsheet and given some comments, and last night I responded to him. Which meant that in the process I was looking over the sheet a great deal and doing lots of mathematical conversions and the like.

Anyway, I went to bed after I sent the e-mail. And at about 12:30 in the morning, I suddenly shot up in bed because I heard the “click” as something slid into place in my brain. Yeah, I did the whole caricature thing of having the light dawn on me :-)

Of course the only problem is that there’s like maybe a dozen people on Earth who would care about my realization, and I don’t know any of them personally. But I figure why not post it into the Internet anyway? So I will.

First, I should note that with the factor field, I’ve mentioned the “spike” that occurs spaced out every 6 digits. Because of this, I wanted to see what prime numbers would look like in base-6 format (using only 0-5 for your digits, just as binary uses only 1 and 0). Last night, I compiled a short list of some of the primes and e-mailed them to the person I’ve been corresponding with, so here’s the list of primes from 2 - 101 with their corresponding base-6 conversion:

2 = 2
3 = 3
5 = 5
7 = 11
11 = 15
13 = 21
17 = 25
19 = 31
23 = 35
29 = 45
31 = 51
37 = 101
41 = 105
43 = 111
47 = 115
53 = 125
59 = 135
61 = 141
67 = 151
71 = 155
73 = 201
79 = 211
83 = 215
89 = 225
97 = 241
101 = 245

And just for fun, converting the last 10 primes on the Excel sheet gives us this:

65413 = 1222501
65419 = 1222511
65423 = 1222515
65437 = 1222541
65447 = 1222555
65449 = 1223001
65479 = 1223051
65497 = 1223121
65519 = 1223155
65521 = 1223201

So as you can see, all the prime numbers after 2 & 3 end in either 1 or 5 in base-6.

Now because my hypothesis (which I lack the mathematical skills to prove beyond a shadow of a doubt) is that all prime numbers end in 1 or 5 in base-6, as I was trying to fall asleep I thought: “At what point do the prime numbers interfere with the 6-spike?” That is, at what point on the number series do prime numbers fall either 1 above or 1 below the spike (1 above the spike corresponds to a number ending in 5, one below corresponds to a number ending in 1).

Obviously 2 and 3 are ruled out from the get-go, because 2 x 3 creates the 6-spike. So I started with 5. And here’s what I got:

In this graphic, the blue lines are the 6-spike. The red cells are those that occur either 1 above or 1 below the 6-spike. The black cells are the other cells that do not fall either one above or one below the 6-spike.

I left the factors in the cells too. As a result, trace the 5-line down and you see that the first time it falls 1 above or 1 below the 6-spike (1 above in this case) is at 5 x 1. The next time it falls 1 above or 1 below the 6-spike (1 below in this case) is at 5 x 5. The next time (1 above) is at 5 x 7. Then again (1 below) at 5 x 11. Finally, it comes 1 above at 5 x 13.

Now look at the 7 line. 7 does the exact same thing but with the above/below polarity switched! The first time it appears is 1 below at 7 x 1. Then at 1 above at 7 x 5, etc. We see the same thing with the 11 and 13 lines. Thus we have:

N = multiple of 6.

N - 1 goes in an above/below sequence.

N + 1 goes in a below/above sequence.

And the real kicker…the N +/- 1 is itself the number that the factors are based on! Thus, take a factor of 6. Subtract 1. It is now 1 below a factor of 6. Multiply by 5 (i.e. 6 -1) and you will be 1 above a factor of 6. Multiply by 7 (i.e. 6 + 1) and you will be 1 below a factor of 6. Multiply by 11 (i.e. [2 x 6] – 1) and you will be one above a factor of 6. Multiply by 13 (i.e. [2 x 6] + 1) and you will be one below a factor of 6.

Let’s give an example. 24 is a factor of 6.

24 – 1 = 23. 23 x 5 = 115. 115 – 1 = 114. 114 = 6 x 19.

24 + 1 = 25. 25 x 5 = 125. 125 + 1 = 126. 126 = 6 x 21.

24 – 1 = 23. 23 x 7 = 161. 161 +1 = 162. 162 = 6 x 27.

24 + 1 = 25. 25 x 7 = 175. 175 – 1 = 174. 174 = 6 x 29.

24 + (2 x 6) – 1 = 35. 35 x 5 = 175. 175 – 1 = 174. 174 = 6 x 29.

24 + (2 x 6) + 1 = 37. 37 x 5 = 185. 185 + 1 = 186. 186 = 6 x 31.

24 + (2 x 6) – 1 = 35. 35 x 7 = 245. 245 + 1 = 246. 246 = 6 x 41.

24 + (2 x 6) + 1 = 37. 37 x 7 = 259. 259 – 1 = 258. 258 = 6 x 43.

So, to generalize it further, let us define an N class number as a positive number that is divisible evenly by 6.

1. (N_x - 1) x (N_y - 1) = X. X – 1 is an N class number.
2. (N_x + 1) x (N_y - 1) = X. X + 1 is an N class number.
3. (N_x - 1) x (N_y + 1) = X. X + 1 is an N class number.
4. (N_x + 1) x (N_y +1) = X. X – 1 is an N class number.

To test this, let N_x = 36 and N_y = 12.

1. (36 – 1) x (12 – 1) = 35 x 11 = 385. Subtract 1 and 384 = 6 x 64.
2. (36 + 1) x (12 -1) = 37 x 11 = 407. Add 1 and 408 = 6 x 68.
3. (36 – 1) x (12 +1) = 35 x 13 = 455. Add 1 and 456 = 6 x 76.
4. (36 + 1) x (12 + 1) = 37 x 13 = 481. Subtract 1 and 480 = 6 x 80.

But we can further generalize this by creating a new class, which I will call the P class. The P class is defined as any number that is N +/- 1. So take any N class, add or subtract one from it, and that is a P class number. From the above, we therefore know that any P class multiplied by another P class number yields another P class number. It comes in the following format.

Let us define P_down as a N – 1 class number and P_up as an N + 1 number.

1. P_down x P_down = P_down.
2. P_up x P_down = P_up.
3. P_down x P_up = P_up.
4. P_up x P_up = P_down.

Now my theory is that all prime numbers are P class numbers, but not all P class numbers are prime numbers. After all, since a P class x a P class yields a P class, then we have proof that P classes can exist with factors. But here’s my theory on that: the only P class numbers that are not primes are those P classes that are created by multiplying other P class variables.

In other words, when thinking about primes, one need not worry about anything other than P class integers.

Let me explain by showing the first few primes again. After 2 and 3 (which create the 6-spike in the first place) we have 5, 7, 11, 13, 17, 19. Each of these shows both sides of the 6-spike.

The first “break” occurs after 23, because 25 has factors. But what are the factors of 25? Only 5 x 5. And 5 is a prime number. In fact, 5 is the smallest prime number that comes into play (again, because 2 and 3 are working to create the 6-spike so they are irrelevant here). In fact, if we multiply the smallest relevant primes, we get:

5 x 5 = 25.
5 x 7 = 35
7 x 7 = 49
5 x 11 = 55
5 x 13 = 65
7 x 11 = 77
5 x 17 = 85
7 x 13 = 91
5 x 19 = 95

And these results are all the numbers that are missing from the 6-spike as primes.

In any case, I think it’s safe to say that we can define a prime number as any P class number that is not divisible by any other P class number. And I also think that P class number that are divisible by any numbers at all are only divisible by other P class numbers. Therefore, we need not worry about any other numbers when testing for primes.

One thing that can be both a blessing and a curse about my nature is that I am often able to find ways to keep myself awake almost all night long. It doesn’t matter how tired I am when I try to go to bed, if I think of something that gets my mind going then I’d much rather continue to think on it than sleep. This happened to me the other day as I was thinking of a few statistical quirks regarding Natural Selection, random mutations, and the like.

Unfortunately, I’m not yet able to write the blog post that I wanted to write, because while I know exactly what I’m talking about, it lacks sufficient groundwork for many other readers to be able to follow along! Since I am an apologist at heart (one who would love to preside over the complete destruction of ideological Darwinism, mind you) I do wish to expand on my thoughts and present them to others, so this leaves me with the necessary task of providing some starting groundwork before I get to the main point. And besides, although it’s tangential to my ultimate point, some of this stuff is just plain kewl :-)

In any case, since a great deal of what I will be focusing on in future posts will deal with statistical analysis, I thought it might be beneficial to give a quick overview of The Three Prisoners Problem in order to A) melt your brain if you’ve never heard of it and B) show how statistics can be logical and yet make no sense at first glance (mostly due to a wrong perspective).

The Three Prisoners Problem was originally mentioned by Martin Gardner in his “Mathematical Games” column in the October, 1959 edition of Scientific American, but under the Monty Hall guise (its mathematical equivalent) it has gathered more infamy, especially after Marilyn vos Savant’s article in Parade magazine in 1990. If this doesn’t make sense to you at first glance, you can take comfort in the fact that it has fooled Nobel laureates, professional mathematicians, and Mensa members countless times. Here I will give my own version of the problem.

There are three prisoners in the king’s dungeon: Adam, Bill, and Charlie. The Warden arrives at each cell and says, “The King has decided that two of you shall go free tomorrow.” At this, there is great rejoicing. But the Warden continues: “However, one of you will be executed.”

“Who will it be?” they all ask in turn.

The Warden responds: “The King has told me who will be executed, but he has also forbidden me telling you who will live and who will die.”

Each of the prisoners accepts this answer except for Charlie. Charlie is a shrewd character and because he knows the Warden is scrupulously honest, he asks: “I know you said that you cannot tell me who will be executed or who will be set free, and therefore you cannot tell me my fate. But will you instead give me one name of one of the other prisoners who will be set free?”

The Warden thinks about this for a moment. “Why would you want to know that?” he ponders. “If I don’t give you a name, you know that you have a 1/3 chance of being executed and a 2/3 chance of going free. If I tell you a name, then you will only have a 1/2 chance of going free! It is better for you if you do not know a name.”

“In that case,” Charlie responds, “why not tell me?”

The Warden relents and says, “Adam will go free tomorrow.”

At this, Charlie sits back and smiles because the Warden has inadvertently told him that it is twice as likely that Bill will be executed as it is that he will be executed…

The reason this is a “problem” is because for most of us we reason the way that the Warden did. Surely telling Charlie that Adam will go free has actually reduced Charlie’s odds of survival, hasn’t it? It used to be 2/3 because it could have been Adam, Bill, or Charlie who would be killed and 2 of them would have lived. But now it’s either Bill or Charlie who will be killed and only one of them would live, and that’s a 1/2 chance, isn’t it?

There are two ways to look at this. First, let’s look at the mathematical rule involved: fractional statistics must together add up to 1.

When the prisoners are first given information, there is a 1/3 chance for each of them that they will be killed. Thus, we have the odds of death being:

Adam = 1/3
Bill = 1/3
Charlie = 1/3

Now when Charlie asks which of the first two prisoners will go free, since the Warden is honest, he tells him that Adam is one who will go free. But this gives no new information to Charlie about whether or not Charlie will die. Charlie’s odds of being killed remain 1/3. However, Adam’s odds of being killed are reduced to 0. He will survive.

If Charlie has a 1/3 shot of dying and Adam has a 0 shot of dying, then because statistics must balance to 1 (it is a certainty that someone will die), this means that Bill’s odds of dying must be 2/3. As a result, Bill is twice as likely to be executed as Charlie.

Of course, this still doesn’t seem right at all! After all, how can telling Charlie that Adam will go free affect Bill’s odds of survival but not affect Charlie’s original odds of survival?

The second way of explaining this helps to flesh it out a bit better. As we stated, when the problem begins, each prisoner has a 1/3 chance of being killed. Therefore, there are three possible options. Let us examine these three options and what the Warden must respond under each option.

Option 1: Adam is killed. If Adam is the one to be executed, then when Charlie asks for the name of one of the two prisoners who will live, the Warden must respond “Bill.” If he says Adam lives, then he has lied (and we’ve stipulated that the Warden is honest). Conclusion: Charlie lives; the prisoner not named dies.

Option 2: Bill is killed. Like the above, the Warden’s choice is restricted to one answer. The Warden can only say that “Adam” will live. Conclusion: Charlie lives; the prisoner not named dies.

Option 3: Charlie is killed. Here is the only instance where the Warden has freedom. Since Charlie will be killed, then he can name either Adam or Bill. Conclusion: Charlie dies; the prisoner not named lives.

As we see in the above, Charlie’s chances of being killed remain 1/3 because only under option 3 does he die. Further, 2/3 of the time the Warden is forced to name a specific prisoner because the one not named is the one who will die. Therefore, 2/3 of the time the prisoner not named is the prisoner who will be executed.

This is also easier to see if we use bigger numbers. Suppose that there are instead 1,000 prisoners and all but one of them will be set free while the remaining prisoner is executed. Under these circumstances, the Warden reveals 998 prisoners who will be set free, leaving only Charlie and prisoner number 473 behind. Which is more likely, that the Warden was forced to leave prisoner number 473 as an option or that Charlie is going to be killed and prisoner 473 was a random selection? Obviously, there is only a 1/1000 chance that prisoner 473 was a random selection, but there is a 999/1000 chance that prisoner 473 was the forced choice. So in this case, the reason it is counterintuitive has more to do with the fact that we do not realize the Warden is excluding all but one prisoner from his answer. If there were 1,000 prisoners total and Charlie asked for the list of 998 of them that would go free, the Warden would immediately spot this error.

Note, however, that even under the circumstance that Charlie only asked for the name of one prisoner out of the 1000 who would go free, that would decrease the odds of all the other unnamed prisoners surviving, although in this instance the amount the odds change would be negligible. Charlie would remain with a 1/1000 chance of dying, while the 998 unnamed prisoners would have just over a 1.001/1000 chance of dying and the one named one would have a no (0) chance of dying. This equates to 998 prisoners splitting a 999/1000 odds, so you still end up with 1/1000 + 999/1000 + 0 = 1. (1.001 x 998 rounds to 999.)

As I mentioned at the top of this post, this is mathematically equivalent to the Monty Hall Problem. That can be demonstrated while keeping with the prisoner motif in the following manner. Suppose that instead of the Warden talking to the prisoners, the King summons the Warden to his throne room. The King, who enjoys tormenting the Warden, says:

“Warden, I am going to execute two prisoners tomorrow, but I am going to free one of them. I have written his name down and locked it in this chest beside me along with one thousand gold pieces. If you can guess who will go free, you can have all the gold in the chest. If you do not guess who goes free, you will have to join the prisoners being executed!”

The Warden realizes he has a 1/3 chance of gaining riches and a 2/3 chance of dying. Nevertheless, the King has given him no option. So he says, “I pick Adam to live.”

The King smiles and says, “Let us make this more interesting. Before you open the chest and see the name, I will tell you that Charlie is going to die. Now, do you still want to choose Adam to live, or do you want to switch your choice to Bill?”

At this point, what should the Warden decide?

Again, mathematically this is equivalent to the Three Prisoners Problem above. Therefore, we know that when the Warden picked Adam to live, he had a 1/3 chance of being right. The King has now informed the Warden that Charlie will die: therefore, Charlie has a 0 chance of living. Once again, because the numbers have to add up to 1, this means that Bill now has a 2/3 chance of living and Adam only has a 1/3 chance of living. Therefore, the Warden should switch his choice.

And to demonstrate this in the similar manner as above, look at the three options of what would happen after the Warden picks Adam but before the King (who already know who will die) responds:

Option 1: Adam lives. In this case, the King can name either Bill or Charlie as dying. Therefore, the Warden should not switch his choice because whomever the King does not name of the other two prisoners will die.

Option 2: Bill lives. In this case, the King MUST name Charlie as dying. The Warden should change his pick to the prisoner not mentioned (Bill).

Option 3: Charlie lives. In this case, the King MUST name Bill as dying. The Warden should change his pick to the prisoner not mentioned (Charlie).

Again we see that 2/3 of the time, the Warden should change his selection.

So here we see that sometimes statistics can be perfectly logical and rational, yet the result is so counterintuitive that they feel wrong. In my next post, I’ll give an example of the opposite: when statistics are irrational and yet seem to make sense. After that, I will look at a few examples statistics in action with Darwinism.