111… has mr = 1 because this is a reducible pattern, reducing to 1…

The latter is an important example because one might be tempted to say this pattern repeats every 3rd, every 3rd offset by 1 and every 3rd offset by 2, but this pattern reduces to 1… therefore, mr is calculated on lowest reducible pattern.

Examining Pat(n) re: Randomness with increasing n So as we take higher n in Pat(n), number of smallest repeating units increases. In fact it exactly equals n. For any given n, Pat(n) isn’t absolutely random, but P(n+1) is more random than Pat(n).

Model of Lim(x->inf) (1/x) = 0 Examine model of:

let f(x) = 1/x

At no x, is f(x) = 0, however

closenesstozero(f(x+1))>closenesstozero(f(x))

and the

lim(x->inf)f(x) = 0

Likewise, for no x is Pat(x) a random pattern, however

The mr(Pat(n+1))>mr(Pat(n)), and then the

Important Identities (4a) lim(n->inf) mr(Pat(n)) = inf (i.e. grows infinitely complex)

(4b) lim(n->inf) uniqueContribution(P(n)) = random set

(4c) lim(n->inf) Pat(n) = random binary pattern (i.e. absolute random)

Definition of Random in English Pat(n) always produces patterns in lowest reducible form (ask me for a proof if you like). Pat(n) has n smallest repeating units (we know this because units are prime). Therefore as you create Pat(n) with greater and greater n, you produce lowest reducible patterns of greater and greater complexity (higher mr). Each individual P(n) as n increases, has a more and more complicated uniqueContribution, leading to more complexity in resulting Pat(n)’s, hence more randomness. As you do this without bound, you create complexity based on previous complexity, resulting in infinite complexity = random.

So what can we do with this knowledge?

Solution to prime twin, triple, quadruplet problem Well it solves prime twin, triplet, and quadruplet problems in a shot...

From 2 above, we know that zero's between P(n) and P(n)^2 are prime. A prime twin will occur in this region whenever you see pattern 00 (two adjacent prime candidates). Can one predictably say that there exists a certain Prime P(K) after which, there will never be a 00 in pattern between P(k+q) and P(k+q)^2?

An examination of pattern combinatorics reveals that there is a 00 in base case P(1) (100..). As we combine patterns, there will always be a 00 somewhere in pattern Pat(n) (ask me for proof if you like). The trick is, will it be between P(n) and P(n)^2.

Well pattern subtended by P(n) and P(n)^2 is a subset of pattern Pat(n) and grows without bound as n does. You can tell me less and less about it as n grows without bound. By (4) I can let n grow without bound until it is a truly random pattern at which time you can no longer tell me that you can predictably state that there won't be a 00 in pattern between P(n) and P(n)^2. At time that there is a 00 between P(n) and P(n)^2, a new prime twin will occur.

This is true of all prime triples, quadruplets etc that are allowable.

Why do they keep finding patterns in primes? This becomes evident knowing that we only have a finite list of primes in our knowledge. The patterns produced by a finite list of prime factors are never absolutely random, just relatively random, or ‘sufficiently complex to avoid simple categorization’. Statistical tools, depending on their power, will find a pattern in those patterns not produced by an infinite number of prime factors (the number line).

Interesting Patterns in Non-Primes There are some interesting patterns in non-primes that emerge from this work.

Define:

LowMarker(n) = 3 + 2(P(1)xP(2)x…x(P(n)),

HighMarker(n) = 3 + 2(3x5x…xP(n)), and

Offset(n) = P(n) – 3

We can say conclusively, thanks to pattern combinatorics that numbers in ranges:

LowRepeater(n) = [LowMarker(n),LowMarker(n)+Offset(n)] and

HighRepeater(n) = [HighMarker(n),HighMarker(n)+Offset(n)] are non-prime (product of primes)

Moreover they follow a similar pattern to base pattern that spawned them.

Example Let’s work an example:

Examine Pat(4) at start of pattern Examine numbers

{3,5,7,9,11}

3 is a product of 3,

5 is a product of 5,

7 is a product of 7,

9 is a product of both 9 and 3,

11 is a product of 11

Recall P(4) = 11

Let’s examine numbers in ranges, LowRepeater(4) and HighRepeater(4)

Examine LowRepeater(4) LowRepeater(4) = {2313,2315,2317,2319,2321}

2313 is a product of 3,

2315 is a product of 5,

2317 is a product of 7,

2319 is a product 3, and necessarily, not a product of 9 (this could only occur in HighRepeater)

2321 is a product of 11

Examine HighRepeater(4) HighRepeater(4) = {20793, 20795, 20797,20799, 20801}

20793 is a product of 3,

20795 is a product of 5,

20797 is a product of 7,

20799 is a product of both 9 and 3,

20801 is a product of 11

LowRepeater(n,k) and HighRepeater(n,k) LowRepeater and HighRepeater repeat over number line

Adding a factor k to previous functions we get:

LowMarker(n,k) = 3 + 2k(P(1)xP(2)x…x(P(n)),

HighMarker(n,k) = 3 + 2k(3x5x…xP(n)), and

Offset(n) = P(n) – 3

Then,

LowRepeater(n,k) = [LowMarker(n,k),LowMarker(n,k)+Offset(n)] and

HighRepeater(n,k) = [HighMarker(n,k),HighMarker(n,k)+Offset(n)] are non-prime (product of primes)

All, where k>=0 and k is an integer.

Interesting observation about difference between LowMarker(n,k) and HighMarker(n,1) For a fixed n, and HighMarker’s k=1, LowMarker’s k = product of non-primes between P(1) and P(n).

The following table will clarify:

LowMarker(3,1) = HighMarker(3,1) LowMarker(4,9) = HighMarker(4,1) LowMarker(5,9) = HighMarker(5,1) LowMarker(6,9*15) = HighMarker(6,1) LowMarker(7,9*15) = HighMarker(7,1) LowMarker(8,9*15*21) = HighMarker(8,1)

Questions? Comments?

martin_winer@hotmail.com

© Martin C. Winer, 2004

Posted to web on Mar 16, 2004 after years of being ignored :(

If you run evaporator blower without running air cond system, sometimes you might get a burst tube in evaporator.

Why? The heat from blower air, evaporating refrigerant in evaporator tubes, and with no where to go (remember air cond compressor is not running), will cause high pressures in tubes.

So be careful that you do not run blower while air cond system is not running.

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