Primes: Randomness and Prime Twin Proof

Written by Martin Winer

Continued from page 1

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,repparttar mr is calculated onrepparttar 127645 lowest reducible pattern.

Examining Pat(n) re: Randomness with increasing n So as we take higher n in Pat(n),repparttar 127646 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 Examinerepparttar 127647 model of:

let f(x) = 1/x

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


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 inrepparttar 127648 lowest reducible form (ask me for a proof if you like). Pat(n) has n smallest repeating units (we know this becauserepparttar 127649 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 inrepparttar 127650 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 solvesrepparttar 127651 prime twin, triplet, and quadruplet problems in a shot...

From 2 above, we know thatrepparttar 127652 zero's between P(n) and P(n)^2 are prime. A prime twin will occur in this region whenever you seerepparttar 127653 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 inrepparttar 127654 pattern between P(k+q) and P(k+q)^2?

An examination of pattern combinatorics reveals that there is a 00 inrepparttar 127655 base case P(1) (100..). As we combine patterns, there will always be a 00 somewhere inrepparttar 127656 pattern Pat(n) (ask me forrepparttar 127657 proof if you like). The trick is, will it be between P(n) and P(n)^2.

Wellrepparttar 127658 pattern subtended by P(n) and P(n)^2 is a subset ofrepparttar 127659 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 inrepparttar 127660 pattern between P(n) and P(n)^2. Atrepparttar 127661 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.


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 inrepparttar 127662 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 torepparttar 127663 base pattern that spawned them.

Example Let’s work an example:

Examine Pat(4) atrepparttar 127664 start ofrepparttar 127665 pattern Examinerepparttar 127666 numbers


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 examinerepparttar 127667 numbers inrepparttar 127668 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 overrepparttar 127669 number line

Adding a factor k torepparttar 127670 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


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 aboutrepparttar 127671 difference between LowMarker(n,k) and HighMarker(n,1) For a fixed n, and HighMarker’s k=1, LowMarker’s k =repparttar 127672 product ofrepparttar 127673 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 C. Winer, 2004

Posted torepparttar 127674 web on Mar 16, 2004 after years of being ignored :(

Martin Winer is a computer scientist by day, working on and an amateur mathematician by night. For a formatted version of this article, please view:

Cooling Down by Evaporating

Written by Thomas Yoon

Continued from page 1

If you runrepparttar evaporator blower without runningrepparttar 127644 air cond system, sometimes you might get a burst tube inrepparttar 127645 evaporator.

Why? The heat fromrepparttar 127646 blower air, evaporatingrepparttar 127647 refrigerant inrepparttar 127648 evaporator tubes, and with no where to go (rememberrepparttar 127649 air cond compressor is not running), will cause high pressures inrepparttar 127650 tubes.

So be careful that you do not runrepparttar 127651 blower whilerepparttar 127652 air cond system is not running.

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