A267028 P(n,k) is an array read by rows, with n > 0 and k=1..5, where row n gives the chain of 5 consecutive primes {p(i), p(i+1), p(i+2), p(i+3), p(i+4)} having the symmetrical property p(i) + p(i+4) = p(i+1) + p(i+3) = 2*p(i+2) for some index i.

18713, 18719, 18731, 18743, 18749, 25603, 25609, 25621, 25633, 25639, 28051, 28057, 28069, 28081, 28087, 30029, 30047, 30059, 30071, 30089, 31033, 31039, 31051, 31063, 31069, 44711, 44729, 44741, 44753, 44771, 76883, 76907, 76913, 76919, 76943

Offset: 1

Author

Michel Lagneau, Feb 23 2016

Keywords

• nonn
• tabf

Comments

a(3 + 5*(n-1)) = A051795(n).

The immediate objective of the sequence is to examine symmetrical properties in the array P(n,k). It is interesting to note that the results with the dimension 5 are generalizable to the dimensions 7, 9, ...

Notation:

We introduce the following function S(i,j) where row i is defined by {P(i,k)} and row j is defined by {P(j,k)}, k = 1..5. Let S(i, j) = 1 if P(i,1) + P(j,5) = P(i,2) + P(j,4) = P(i,3) + P(j,3), otherwise 0.

Conjecture:

For each integer n, there exists an infinite sequence of integers b(n,m), m = 1, 2, ... such that S(n, b(n,m)) = 1.

The following table gives the first values b(n,m).

Notation in the table: "PS" = primitive sequence.

+----+------------------------------------------------+-----------+

| n | sequences b(n,m), m=1,2,... of index |included in|

+----+------------------------------------------------+-----------+

| 1 | 1, 2, 3, 5, 8, 9, 10, 12, 15, 16, 17, 18, ... | PS |

| 2 | 2, 3, 5, 8, 9, 10, 12, 15, 16, 17, 18, 19, ...| {b(1,m)} |

| 3 | 3, 5, 8, 9, 10, 12, 15, 16, 17, 18, 19, ... | {b(1,m)} |

| 4 | 4, 6, 11, 13, 14, 21, 28, 35, 39, 57, 59, ... | PS |

| 5 | 5, 8, 9, 10, 12, 15, 16, 17, 18, 19, 22, ... | {b(1,m)} |

| 6 | 6, 11, 13, 14, 21, 35, 39, 57, 59, 63, 67, ...| {b(4,m)} |

| 7 | 7, 30, 52, 55, 73, 74, 115, 159, 177, 183, ...| PS |

| 8 | 8, 9, 10, 12, 15, 16, 17, 18, 19, 22, 23, ... | {b(1,m)} |

| 9 | 9, 10, 12, 15, 16, 17, 18, 19, 22, 23, 24, ...| {b(1,m)} |

| 10 | 10, 12, 15, 16, 17, 18, 19, 22, 23, 24, 26, ...| {b(1,m)} |

| 11 | 11, 13, 14, 21, 28, 35, 39, 57, 59, 63, 67, ...| {b(4,m)} |

| 12 | 12, 15, 16, 17, 18, 19, 22, 23, 24, 26, 27, ...| {b(1,m)} |

| 13 | 13, 14, 21, 28, 35, 39, 57, 59, 63, 67, 70, ...| {b(4,m)} |

| .. | ... | ... |

| 20 | 20, 43, 56, 96, 113, 131, 135, 156, 196, ... | PS |

| 25 | 21, 33, 37, 38, 40, 47, 48, 65, 76, 79, 83, ...| PS |

...

Example: S(7, 30) = 1.

We observe primitive sequences {b(n,m)} for n = {1, 4, 7, 20, 25, ...}.

(A primitive sequence is a sequence which is not included in another.)

Properties:

(1) S(i, i)= 1 for all i;

(2) S(i, j) = 1 => S(j, i) = 1;

(3) S(i, j) = 1 and S(j, L) = 1 => S(i, L) = 1.

Example:

For n = 1, {P(1,k)} = {18713, 18719, 18731, 18743, 18749};

we choose, for instance, b(1,2) = 3 => for n = 3, {C(3,k)} = {28051, 28057, 28069, 28081, 28087};

S(1,3) = 1 because 18713 + 28087 = 18719 + 28081 = 18731 + 28069 = 18743 + 28057 = 18749 + 28051 = 46800.

In order to find the index L for satisfying the property (3), we choose, for instance, the index b(3,2) = 8 => for n = 8, {P(8,k)} = {97423, 97429, 97441, 97453, 97459} and S(3, 8) = 1 because 28051 + 97459 = 28057 + 97453 = 28069 + 97441 = 28081 + 97429 = 28087 + 97423 = 125510.

Conclusion: S(1, 3) = 1 and S(3, 8) = 1 => S(1, 8) = 1 with 18713 + 97459 = 18719 + 97453 = 18731 + 97441 = 18743 + 97429 = 18749 + 97423 = 116172.

Examples

			The first row is [18713, 18719, 18731, 18743, 18749] because 18713 + 18749 = 18719 + 18743 = 2*18731 = 37462.
The array starts with:
  [18713, 18719, 18731, 18743, 18749]
  [25603, 25609, 25621, 25633, 25639]
  [28051, 28057, 28069, 28081, 28087]
  ...
		

Crossrefs

Cf. A051795, A055380, A055382.

Programs

Maple

U:=array(1..50,1..5):W:=array(1..2):kk:=0:
for n from 4 to 10000 do:
   for m from 2 by -1 to 1 do:
      q:=ithprime(n-m)+ithprime(n+m):W[m]:=q:
    od:
    if W[1]=W[2] and W[1]=2*ithprime(n) then
    kk:=kk+1:U[kk,1]:=ithprime(n-2):
    U[kk,2]:=ithprime(n-1):U[kk,3]:=ithprime(n):
    U[kk,4]:=ithprime(n+1):U[kk,5]:=ithprime(n+2):
    else fi:od:print(U):
    for i from 1 to kk do:
     for j from i+1 to kk do:
      s1:=U[i,1]+U[j,5]:
      s2:=U[i,2]+U[j,4]:
      s3:=U[i,3]+U[j,3]:
      s4:=U[i,4]+U[j,2]:
      s5:=U[i,5]+U[j,1]:
     if s1=s2 and s2=s3 and s3=s4 and s4=s5
     then
     printf("%d %d \n",i,j):
     else fi:
     od:
  od: