Prime numbers and primality testing
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Gaps Between Consecutive Odds not Divisible by 3

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w_sindelar@juno.com     Message 1 of 5  Jun 27, 2003
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Let "N" represent an ODD COMPOSITE INTEGER NOT DIVISIBLE by 3. 
Let A < B represent a pair of CONSECUTIVE N's, meaning that there is no N
greater than A and less than B. 
For example, here is a short list of the first few consecutive N's; 25,
35, 49, 55, 65, 77, 85. Any 2 adjacent integers A and B in the list are
called a pair of consecutive odd composite integers not divisible by 3.
Let G= (B - A) call it the GAP. 
The hunt for huge gaps between consecutive PRIMES seems to imply that
there is no limit to the size of the gap. As a lark I thought I would see
how large of a gap "G" I could find between the 2 consecutive composites
"A" and "B" and how many primes exist between them. The largest gap I
could find was 20 with 6 primes between!! Can it be possible that is the
largest gap possible? Would anyone care to have a go at this with a
program other than Ubasic? 
Thanks folks and regards.
Bill Sindelar
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Mark Underwood     Message 2 of 5  Jun 27, 2003
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Hi Bill et all

The composite N's have factors of the form 6t+1 or 6t-1.

The largest gap possible would be of the form
5*(6*(t+1) -1) - 5*(6t +1)

which is 20, which you have found. 

Speaking of multiples of 5, while gazing at a sheet of primes the 
other day (hehe) I noticed four consecutive primes ending in the same 
number: 22963, 22973, 22993 and 23003. If consecutive primes were 
independant events then I guess the odds of four consecutive primes 
ending in the same number in the set of primes would be 
4 * 1/4 * 1/4 * 1/4 * 1/4  
which is 1/64. Yet on inspection four consecutive primes of this 
nature appear to be rarer than this.  Perhaps up higher in the list 
of primes it approaches more this probability.

For the fun of it, can anyone produce a longer string of such primes?

Mark

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Mikael Klasson     Message 3 of 5  Jun 27, 2003
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Hi,

> Speaking of multiples of 5, while gazing at a sheet of primes the
> other day (hehe) I noticed four consecutive primes ending in the same
> number: 22963, 22973, 22993 and 23003. If consecutive primes were
> independant events then I guess the odds of four consecutive primes
> ending in the same number in the set of primes would be
> 4 * 1/4 * 1/4 * 1/4 * 1/4
> which is 1/64. Yet on inspection four consecutive primes of this
> nature appear to be rarer than this.  Perhaps up higher in the list
> of primes it approaches more this probability.
> 
> For the fun of it, can anyone produce a longer string of such primes?

For our mutual amusement, I humbly present:
452942837,452942857,452942927,452942947,452942977,452943047,452943097,452943107,
452943137,452943157
:)

Mikael
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Mark Underwood     Message 4 of 5  Jun 28, 2003
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Hi all

Phil Carmody has generated two data sets, each tallying the occurance 
of consecutive primes ending in the same number.

In the first data set we have

Length of consecutive     Number of Occurances   
prime string

7                         112783      
8                         17612  
9                         2745 
10                        438
11                        59
12                        7
13                        0
14                        1

In the second data set we have

Length of consecutive     Number of Occurances   
prime string

11                        2323
12                        360
13                        62
14                        10
15                        2

What surprised me was that the number of occurances does not decrease 
by a factor of 4 with each unit increase of string length. Rather it 
appears to decrease by a factor of roughly 6.4 when the occurances 
are high enough to stabilize the ratio.

If anyone would like to try their hand at explaining why it is about 
6.4 rather than 4, I would like to hear it! It appears that 
consecutive primes separated by a multiple of 10 deviate considerably 
from being considered independent events, but that's about all I can 
conjure up myself right now.

Phil has a graph of the first occurances of these prime strings at
http://fatphil.org/maths/trivia/terminal.html

Mark
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Mark Underwood     Message 5 of 5  Jun 29, 2003
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Hi all,

I thought I should clarify somethings about the last post which were 
probably confusing. 

Phil Carmody has compiled two data sets from two different test 
intervals. The first interval went from the prime 3,873,011 to the 
prime 159,163,479,499.  

In this first interval 

Strings of exactly length 7 occur 112783 times.
Strings of exactly length 8 occur 17612 times.
Strings of exactly length 9 occur 2745 times.
Strings of exactly length 10 occur 438 times.
Strings of exactly length 11 occur 59 times.
Strings of exactly length 12 occur 7 times.
Strings of exactly length 13 occur 0 times.
Strings of exactly length 14 occur 1 times.

The second test interval went from the prime 34,239,812,903 to the 
prime 4,053,462,479,317 and hence overlaps the first test interval.  

In this second interval

Strings of exactly length 11 occur 2323 times.
Strings of exactly length 12 occur 360 times.
Strings of exactly length 13 occur 62 times.
Strings of exactly length 14 occur 10 times.
Strings of exactly length 15 occur 2 times.

What surprised me at first was that the number of occurances does not 
decrease by a factor of 4 with each unit increase of string length. 
Since primes end in only four digits - 1,3,7 and 9 - if these digits 
were distributed randomly and equally one would expect that there 
would be a one in four chance that two consecutive primes would end 
in the same number.  Similarly one would expect strings of n 
consecutive primes ending in the same digit to be 4 times as abundant 
as strings of n+1 consecutive primes ending in the same digit.

Yet Phil's data shows that strings of n consecutive primes ending in 
the same digit appear to be about 6.4 times as abundant as strings of 
n+1 consecutive primes ending in the same digit. That is, when 
occurances are large enough to stabilize the ratio.

I suppose that the progression of the average size of prime gaps in 
an interval may play a part, and moreso the fact that certain 
spacings between primes are more likely.  But can anyone produce the 
details of how the 6.4 ratio comes about?

Mark


PS 

In addition to this, and for interests sake, Phil Carmody has 
compiled a listing of the first occurences of these prime strings.

At  http://fatphil.org/maths/trivia/terminal.html  Phil's uses the 
convention a(n) to represent the first prime of the first string to 
contain exactly n consecutive primes having the same last digit.

For instance, since 139,149 are the first two consecutive primes that 
have the same last digit,  a(2) = 139.

Since 1627,1637 and 1657 are the first three consecutive primes that 
have the same last digit, a(3) = 1627.

Here is Phil's compilation: 

a(1) =  2
a(2) =  139
a(3) =  1,627
a(4) =  18,839
a(5) =  123,229
a(6) =  776,257
a(7) =  3,873,011
a(8) =  23,884,639
a(9) =  36,539,311
a(10) = 196,943,081
a(11) = 452,942,827
a(12) = 73,712,513,057
a(13) = 177,746,482,483
a(14) = 154,351,758,091
a(15) = 4,010,803,176,619
a(16) = ???
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