cp's OEIS Frontend

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.

%I A267028 #68 Nov 02 2021 22:24:51
%S A267028 18713,18719,18731,18743,18749,25603,25609,25621,25633,25639,28051,
%T A267028 28057,28069,28081,28087,30029,30047,30059,30071,30089,31033,31039,
%U A267028 31051,31063,31069,44711,44729,44741,44753,44771,76883,76907,76913,76919,76943
%N 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.
%C A267028 a(3 + 5*(n-1)) = A051795(n).
%C A267028 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, ...
%C A267028 Notation:
%C A267028 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.
%C A267028 Conjecture:
%C A267028 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.
%C A267028 The following table gives the first values b(n,m).
%C A267028 Notation in the table: "PS" = primitive sequence.
%C A267028 +----+------------------------------------------------+-----------+
%C A267028 |  n |      sequences b(n,m), m=1,2,... of index      |included in|
%C A267028 +----+------------------------------------------------+-----------+
%C A267028 |  1 |  1, 2, 3, 5, 8, 9, 10, 12, 15, 16, 17, 18, ... |     PS    |
%C A267028 |  2 |  2, 3, 5, 8, 9, 10, 12, 15, 16, 17, 18, 19, ...|  {b(1,m)} |
%C A267028 |  3 |  3, 5, 8, 9, 10, 12, 15, 16, 17, 18, 19, ...   |  {b(1,m)} |
%C A267028 |  4 |  4, 6, 11, 13, 14, 21, 28, 35, 39, 57, 59, ... |     PS    |
%C A267028 |  5 |  5, 8, 9, 10, 12, 15, 16, 17, 18, 19, 22, ...  |  {b(1,m)} |
%C A267028 |  6 |  6, 11, 13, 14, 21, 35, 39, 57, 59, 63, 67, ...|  {b(4,m)} |
%C A267028 |  7 |  7, 30, 52, 55, 73, 74, 115, 159, 177, 183, ...|     PS    |
%C A267028 |  8 |  8, 9, 10, 12, 15, 16, 17, 18, 19, 22, 23, ... |  {b(1,m)} |
%C A267028 |  9 |  9, 10, 12, 15, 16, 17, 18, 19, 22, 23, 24, ...|  {b(1,m)} |
%C A267028 | 10 | 10, 12, 15, 16, 17, 18, 19, 22, 23, 24, 26, ...|  {b(1,m)} |
%C A267028 | 11 | 11, 13, 14, 21, 28, 35, 39, 57, 59, 63, 67, ...|  {b(4,m)} |
%C A267028 | 12 | 12, 15, 16, 17, 18, 19, 22, 23, 24, 26, 27, ...|  {b(1,m)} |
%C A267028 | 13 | 13, 14, 21, 28, 35, 39, 57, 59, 63, 67, 70, ...|  {b(4,m)} |
%C A267028 | .. | ...                                            |     ...   |
%C A267028 | 20 | 20, 43, 56, 96, 113, 131, 135, 156, 196, ...   |     PS    |
%C A267028 | 25 | 21, 33, 37, 38, 40, 47, 48, 65, 76, 79, 83, ...|     PS    |
%C A267028 ...
%C A267028 Example: S(7, 30) = 1.
%C A267028 We observe primitive sequences {b(n,m)} for n = {1, 4, 7, 20, 25, ...}.
%C A267028 (A primitive sequence is a sequence which is not included in another.)
%C A267028 Properties:
%C A267028 (1) S(i, i)= 1 for all i;
%C A267028 (2) S(i, j) = 1 => S(j, i) = 1;
%C A267028 (3) S(i, j) = 1 and S(j, L) = 1 => S(i, L) = 1.
%C A267028 Example:
%C A267028 For n = 1, {P(1,k)} = {18713, 18719, 18731, 18743, 18749};
%C A267028 we choose, for instance, b(1,2) = 3 => for n = 3, {C(3,k)} = {28051, 28057, 28069, 28081, 28087};
%C A267028 S(1,3) = 1 because 18713 + 28087 = 18719 + 28081 = 18731 + 28069 = 18743 + 28057 = 18749 + 28051 = 46800.
%C A267028 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.
%C A267028 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.
%e A267028 The first row is [18713, 18719, 18731, 18743, 18749] because 18713 + 18749 = 18719 + 18743 = 2*18731 = 37462.
%e A267028 The array starts with:
%e A267028   [18713, 18719, 18731, 18743, 18749]
%e A267028   [25603, 25609, 25621, 25633, 25639]
%e A267028   [28051, 28057, 28069, 28081, 28087]
%e A267028   ...
%p A267028 U:=array(1..50,1..5):W:=array(1..2):kk:=0:
%p A267028 for n from 4 to 10000 do:
%p A267028    for m from 2 by -1 to 1 do:
%p A267028       q:=ithprime(n-m)+ithprime(n+m):W[m]:=q:
%p A267028     od:
%p A267028     if W[1]=W[2] and W[1]=2*ithprime(n) then
%p A267028     kk:=kk+1:U[kk,1]:=ithprime(n-2):
%p A267028     U[kk,2]:=ithprime(n-1):U[kk,3]:=ithprime(n):
%p A267028     U[kk,4]:=ithprime(n+1):U[kk,5]:=ithprime(n+2):
%p A267028     else fi:od:print(U):
%p A267028     for i from 1 to kk do:
%p A267028      for j from i+1 to kk do:
%p A267028       s1:=U[i,1]+U[j,5]:
%p A267028       s2:=U[i,2]+U[j,4]:
%p A267028       s3:=U[i,3]+U[j,3]:
%p A267028       s4:=U[i,4]+U[j,2]:
%p A267028       s5:=U[i,5]+U[j,1]:
%p A267028      if s1=s2 and s2=s3 and s3=s4 and s4=s5
%p A267028      then
%p A267028      printf("%d %d \n",i,j):
%p A267028      else fi:
%p A267028      od:
%p A267028   od:
%Y A267028 Cf. A051795, A055380, A055382.
%K A267028 nonn,tabf
%O A267028 1,1
%A A267028 _Michel Lagneau_, Feb 23 2016