Problems & Puzzles: Puzzles

Problem 73. Minimal symmetric tuples from consecutive twin primes...

Natalia Makarova sent the following interesting puzzle;


Definition

the tuples from k consecutive twin primes

p1, p1 + 2, p2, p2 + 2, ..., pk-1, pk-1 + 2, pk, pk + 2

is symmetric if the following condition is satisfied

p1 + pk + 2 = p1 + 2 + pk = p2 + pk-1 + 2 = p2 + 2 + pk-1 = ...

 

Diameter of tuple is the quantity d = pk + 2 - p1.

 

We will consider minimal symmetric tuples from k consecutive twin primes with minimal diameter.

 

d=pk+2-p1

 

k = 2

5, 7, 11, 13

We will write it down briefly
5: 0 2 6 8
d = 8

0 2 6 8 is the pattern of tuple.

 

k = 3

5: 0 2 6 8 12 14

d = 14

 

k = 4

663569: 0 2 12 14 18 20 30 32

d = 32

 

*k = 5

39713433671: 0 2 6 8 18 20 30 32 36 38

d = 38


**k = 6

5008751356547: 0 2 12 14 24 26 30 32 42 44 54 56

d = 56


*k = 7

2485390773085247: 0 2 24 26 30 32 42 44 54 56 60 62 84 86

d = 86

 

I found theoretical patterns for k = 8Ė13.


k = 8

0 2 30 32 42 44 54 56 60 62 72 74 84 86 114 116

d = 116


k = 9

0 2 18 20 30 32 42 44 60 62 78 80 90 92 102 104 120 122

d = 122


k = 10

0 2 12 14 30 32 42 44 54 56 90 92 102 104 114 116 132 134 144 146

0 2 12 14 42 44 54 56 60 62 84 86 90 92 102 104 132 134 144 146

d = 146


k = 11

0 2 12 14 30 32 42 44 54 56 72 74 90 92 102 104 114 116 132 134 144 146

d = 146


k = 12

0 2 30 32 42 44 60 62 72 74 84 86 120 122 132 134 144 146 162 164 174 176 204 206

d = 206


k = 13

0 2 6 8 36 38 48 50 66 68 78 80 108 110 138 140 150 152 168 170 180 182 210 212 216 218

0 2 6 8 36 38 48 50 66 68 90 92 108 110 126 128 150 152 168 170 180 182 210 212 216 218

0 2 6 8 36 38 66 68 78 80 90 92 108 110 126 128 138 140 150 152 180 182 210 212 216 218

0 2 36 38 48 50 66 68 78 80 90 92 108 110 126 128 138 140 150 152 168 170 180 182 216 218

d = 218

 

Q. find minimal symmetric tuples from k consecutive twin primes with minimal diameter for k > 7.

 

_____________

* The solution found in T. Brada Experimental Grid BOINC project

https://boinc.tbrada.eu/

 

** The solution contained in the OEIS sequence

https://oeis.org/A330278

 

Jan van Delden wrote on March 27, 2020:

Iím busy with this problem. Currently testing k=8 and k=12.

I wrote a routine to compute the associated Hardy Littlewood constants.

 

This routine broke down at k=3, for a good reason.

 

The displayed solution:

k = 3

5: 0 2 6 8 12 14

d = 14

 

Is fine. However the pattern [0,2,6,8,12,14] is not in general! The displayed pattern only works with p=5.
If p>5 then all residues modulo 5 are taken: [0,2,1,3,2,4] and we canít choose p. (Except 5 that is).

***


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