Problems & Puzzles:
Problems
Problem
47 . Holes and
Crowds-II
Now we'll focus on the opposite concept to
prime-holes, that is to say, we will discuss these regions where k primes
are as close (crowds) as they can
be,
according to divisibility rules.
This is exactly the core of an issue technically named
prime k-tuples(1).
The best site in the Web to find all kind of theory and records about,
is the well known Tony Forbes's site (patterns,
minimal,
maximal)
Let's suppose we define a k value. Let's suppose, just
to illustrate, that this k value is 8. What things we would be interested to know
about these prime 8-tuplets (p1, p2, ...p8)?
Here are some basic things:
1. What is the minimal distance between the extreme
primes, p1 & p8,
namely, s(8)= p8-p1?.
2. Are there different valid arrangements ( 'patterns') that these 8 primes may
adopt, preserving the same minimal distance s(8)?
3. What is the earliest set of 8 primes, for each pattern?
4. What is the largest known set of 8 primes, for each pattern?
For the case of k=8 the answers are:
1. s(8) = p8 -p1 = 26
2. There are 3 patterns that the eight primes in a 8-tuple may adopt. These
patterns are described by the following set of numbers, representing the
distance between each prime to the first one:
a) pi- p1 = {0 2 6 8 12 18 20 26}
b) pi- p1 = {0 2 6 12 14 20 24 26}
c) pi- p1 = {0 6 8 14 18 20 24 26}
3. The first prime for the earliest example for each pattern is:
p1 = 11, for the first pattern
p1 = 17, for the second pattern
p1 = 88793, for the third
pattern
4. The first prime for the largest known example (June 2003), for each pattern is:
p1= 15234072433401 * 375# + 43813839521, for the first pattern
p1= 243551752728*320# + 1277, for the second pattern
p1 = 22 * 10^38 + 2241278889512323, for the third pattern
Well, Frank Ellerman has noticed that we know
at least one example for at least one pattern, for every 2<=k<=25,
except for k=24
(2)
For k=24, we certainly know that p24-p1 = 100
and that the 24-tuplets may adopt one of four (4) patterns:
a) pi - p1 = {0 4 6 10 12
16 24 30 34 40 42 46 52 60
66 70 72 76 82 84 90 94 96
100}
b) pi - p1 = {0 4 6 12 16 24 30 34
40 42 46 52 54 60 66 70 72
76 82 84 90 94 96 100}
c) pi - p1 = {0 4 6 10 16 18 24 28
30 34 40 46 48 54 58 60 66
70 76 84 88 94 96 100}
d) pi - p1 = {0 4 6 10 16 18 24 28
30 34 40 48 54 58 60 66 70
76 84 88 90 94 96 100}
But the first specific case for k=24, is still waiting its
discoverer. So, the obvious question (also proposed by Ellerman) is:
Q1. Find the earliest p1
such that p24 - p1
=100?
I would like to add one more question:
Q2. Find s(k) and the
quantity of patterns for k=200
_________
(1) Tony Forbes provides the
following rigorous definition of prime k-tuples:
A prime k-tuples is then defined as
a sequence of consecutive primes {p1, p2, ..., pk} such
that for every prime q, not all the residues modulo q are
represented by p1, p2, ..., pk, and pk - p1
= s(k). Observe that the definition excludes a finite number
(for each k) of dense clusters at the beginning of the prime number
sequence - for example, {97, 101, 103, 107, 109} satisfies the conditions of
the definition of a prime 5-tuplet , but {3, 5, 7, 11, 13} doesn't because
all three residues modulo 3 are represented
(2)
Table that summarizes
the valid patterns of k-tuples for 2 to 50
(in red if
no one single example is known, Jun 2001).
s(k) = pk - p1
k |
s(k) |
# of patterns |
2 |
2 |
1 |
3 |
6 |
2 |
4 |
8 |
1 |
5 |
12 |
2 |
6 |
16 |
1 |
7 |
20 |
2 |
8 |
26 |
3 |
9 |
30 |
4 |
10 |
32 |
2 |
11 |
36 |
2 |
12 |
42 |
2 |
13 |
48 |
6 |
14 |
50 |
2 |
15 |
56 |
4 |
16 |
60 |
2 |
17 |
66 |
4 |
18 |
70 |
2 |
19 |
76 |
4 |
20 |
80 |
2 |
21 |
84 |
2 |
22 |
90 |
4 |
23 |
94 |
2 |
24 |
100 |
4 |
25 |
110 |
18 |
26 |
114 |
2 |
27 |
120 |
8 |
28 |
126 |
10 |
29 |
130 |
2 |
30 |
136 |
2 |
31 |
140 |
2 |
32 |
146 |
4 |
33 |
152 |
14 |
34 |
156 |
20 |
35 |
158 |
2 |
36 |
162 |
2 |
37 |
168 |
2 |
38 |
176 |
6 |
39 |
182 |
26 |
40 |
186 |
26 |
41 |
188 |
8 |
42 |
196 |
2 |
43 |
200 |
6 |
44 |
210 |
18 |
45 |
212 |
4 |
46 |
216 |
4 |
47 |
226 |
4 |
48 |
236 |
2 |
49 |
240 |
2 |
50 |
246 |
22 |

Thomas
J. Engelsma has let me know (Set. 2003) that
he has a page
related to these issues and certainly a published
solution for Q2.
***
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