packing density of jerblets question

Also if you want, you can use "fake" primes and put them wherever you want, to see for a given number of fake primes if you can get a jerblet packing density that is higher than the actual placement of primes.

ie can you find 25 numbers below 100 to replace the 25 real primes that can pack in more jerblets than the real primes do?

real primes, how many jerblets?

fake primes, how many jerblets?!

x,y,z,.... (25 of them up to 100)

cheers, Jamie

Ok I don't want anyone's brains to melt so here is a hint for finding jerblets from 1 to 10:

there is only one jerblet using the real prime numbers in that range.

jerblet 3 7

It is a jerblet because 3 and 7 are both prime, and (3+7)/2 is prime (5).

There are four real primes 2,3,5,7 up to 10, so can you find 4 FAKE primes that can produce more than one jerblet?!

cheers, Jamie

BIG MISTAKE WARNING:

I meant to say:

It is a jerblet because 3 and 7 are both prime, and (7-3)/2 is prime (2).

not this: (3+7)/2 is prime (5) (that's incorrect)

A simple case where four fake primes have more jerblets in the range 1 to 10 is for the fake primes 1,2,3,4

There are two jerblets then:

jerblet1: 1 3

jerblet2: 2 4

Any better solutions with more than 2 jerblets?!

cheers, Jamie

Everyone knows that 1 is really a heavily censored prime number, so I have decided to modify the whole question to consider the case of 5 primes from 1 to 10, (or 5 fake primes)

So for the 5 real primes 1,2,3,5,7 these produce a new amazing quantity of SIX jerblets:

jerblet1: 1 3 jerblet2: 1 5 jerblet3: 1 7 jerblet4: 3 5 jerblet5: 3 7 jerblet6: 5 7

proof:

jerblet1: (3-1)/2 = 1 (1 is the new prime) jerblet2: (5-1)/2 = 2 (2 is prime) jerblet3: (7-1)/2 = 3 (3 is prime) jerblet4: (5-3)/2 = 1 (1 is the new prime) jerblet5: (7-3)/2 = 2 (2 is prime) jerblet6: (7-5)/2 = 1 (1 is the new prime)

That is a total of 6 jerblets when 1 is considered a prime number. So go ahead beat that with 5 fake primes from 1 to 10, reward is as many jerblets as you can find.

If you don't like the idea of 1 being considered as a prime, then see if the 25 primes from 1 to 100 have the most jerblets compared to any selection of 25 fake primes.

cheers, Jamie

If you weren't so afraid of having a conversation with yourself you wouldn't be so quick to spread your hatred.

An update after some computation:

I did 1 million runs of random numbers. Each run generated 26 random numbers from 1 to 100, then for each set of the 26 random numbers I checked how many jerblets they had.

I also did a benchmark run to compare the jerblet counts with the 25 primes between 1 and 100, and additionally the number 1, so 26 numbers:

The primes sequence (with 1 added) have 99 jerblets, and the best random sequence that was generated had 89 jerblets, here it is:

So the record at this sequence length has primes at the top of the list still. I think it is likely that with more than 1 million runs, that the random sequence will generate more jerblets than the sequence of primes, but not by much.

My theory is that the primes produce a NEAR OPTIMAL maximum jerblet count for any given number domain ie 1 to 100 in this example, however I think it is possible that there is always the potential to have slightly more jerblets, since the primes (in my opinion) are distributed in a way that takes into account the infinite number domain, so they are optimally distributed over that domain of numbers, and thus only near optimally distributed for shorter number domains. That is just a theory though and to verify it numerically I can run all the possible permutations of random numbers to make sure what the maximum jerblet counts are.

Also for these randomly generated sequences that generate a lot of jerblets, I would like to graph the Z(count) on the y-axis for their Z(x y) primal pair centerpoints to look at the density of x y centerpoints over the range of numbers and compare it to the known Z(count) centerpoint density of the primes which has proportional peaks aligned with the primorial numbers like in these graphs:

zoomed in:

zoomed out:

cheers, Jamie

Restating the original algorithm more generally:

In a range of n numbers using a set of ?(n) numbers called Y, select the set of numbers Y to maximize the count of pairs of numbers x and y from the set Y where x and y are both prime, and where y is larger than x, and with (y-x) / 2 equal to a number that exists in the set Y.

Restating the original algorithm more generally: CORRECTED

In a range of n numbers using a set of ?(n)+1 numbers called Y, select the set of numbers Y to maximize the count of pairs of numbers x and y from the set Y where x and y both exist in Y, and where y is larger than x, and with (y-x) / 2 equal to a number that exists in the set Y.

I hate to be mean, but hasn't this turned into an obsession for you? Would you not be happier on a math forum?

Steve

It's like anything else if you want to figure something out you gotta keep working at it. On a math forum people would reply and guide my search based on their own understanding of math which doesn't include what I'm trying to figure out. Yep big dreams but gotta keep believing :D

Here on the electronics newsgroup I prefer not having to deal with people who can understand the math as it gives me freedom to figure it out on my own.

cheers, Jamie

It was a simple observation of a comical situation - no hatred involved.

But you've got comprehension problems, so it's pointless to try and imagine what you perceived.

Considerate people don't expose their preliminary thoughts until they've worked out some sort of compact and self-consistent story.

Why? At some point you are going to have to come to term with the fact that you are a half-wit.

Sadly, you lack to the capacity to figure your way out of a paper-bag.

Also in addition to the above, all the numbers in Y have to be unique.

More results of computing jerblets (pairs of numbers from the above definition):

range of n numbers = 40 ?(n)+1 = 13

For the case of Y being the primes+1:

jerblet count = 34

10million random runs results for maximum jerblet count: Y = 1,3,4,5,6,7,9,11,13,15,17,27,29 jerblet count = 41

jerblet average for all 10million sequences = 12.259

number of sequences with jerblet counts greater or equal to 34 (the prime sequence) = 373

The prime sequence+1 is in the top 0.00373% of sequences based on jerblet count, I think that general pattern will hold where prime sequences are near the maximum jerblet density.

Also I noticed that the average jerblet counts of all the sequences multipled by e ~= 2.71828 is pretty close to the prime sequence jerblet count. ie 12.259*2.71828=33.323 and the prime jerblet count is 34.

If there is a relationship for calculating the prime sequence jerblet count that would be useful, also if there is a formula to replace the prime counting function ?(n)+1 that would be good too.

cheers, Jamie

Ohhh, it's a spelling mistake, now I get it.. and much more in keeping with SED topics...

Some more data showing the pattern of prime sequences being near the high end of jerblet density:

range of n numbers = 100 ?(n)+1 = 26

For the case of Y being the primes+1:

jerblet count = 99

jerblet count = 109

jerblet average for all 10million sequences = 41.462

number of sequences with jerblet counts greater or equal to 99 (the prime sequence) = 22

The prime sequence+1 is in the top 0.00022% of sequences based on jerblet count, after doing 10million random runs.

cheers, Jamie

uniform random is a bad choice, jerblets favor domains with density at the bottom.