A277970 Numbers k = A005574(m) such that k = (A005574(m-1)+A005574(m+1))/2.

4, 10, 20, 120, 130, 180, 230, 260, 440, 470, 680, 700, 750, 920, 1060, 1320, 1736, 1860, 1970, 2106, 2320, 2460, 2760, 2850, 2890, 3074, 3660, 3800, 4180, 4370, 5030, 5236, 5304, 5814, 5990, 6130, 6350, 6590, 6724, 6780, 6990, 7190, 7384, 7520, 7744, 8180

Offset: 1

Author

Michel Lagneau, Nov 07 2016

Keywords

• nonn

Comments

Subsequence of A005574.

a(n) == 0, 4 or 6 (mod 10).

The corresponding primes a(n)^2+1 are 17, 101, 401, 14401, 16901, ...

The numbers j are symmetric centers given by the middle j of each triple of integers (i, j, k) = (2, 4, 6), (6, 10, 14), (16, 20, 24), (116, 120, 124), ... which are elements of A005574. This symmetry can be seen from the differences between the numbers of each triple. From these, we obtain the following differences (2, 2), (4, 4), (4, 4), (4, 4), .... More generally, a symmetric center may also be the middle of a m-tuple of m even integers (i(1), i(2), ..., i(m)) with m odd, where i(1)^2+1, i(2)^2+1, ..., i(m)^2+1 are m consecutive primes. In order to obtain the symmetry, there must be (i(1)+i(m))/2 = (i(2)+i(m-1))/2 = ... = (i((m-1)/2)+i((m+3)/2))/2 = i((m+1)/2), the middle of the m-tuple.

Because an m-tuple is not unique for each a(n), we introduce the notion of order O(a(n)) (see the table below). The calculations show that O(a(n)) < = 4 for n < 500000.

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

| a(n) |order| m-tuples | differences |

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

| 4 | 1 | (2,4,6) |(2, 2) |

| 10 | 2 | (6,10,14) |(4, 4) |

| | | (4,6,10,14,16) |(2,4,4,2) |

| 20 | 1 | (16,20,24) |(4,4) |

| 120 | 1 | (116,120,124) |(4,4) |

| 130 | 1 | (126,130,134) |(4,4) |

| 180 | 1 | (176,180,184) |(4,4) |

| 230 | 1 | (224,230,236) |(6,6) |

| 260 | 4 | (256,260,264) |(4,4) |

| | | (250,256,260,264,270) |(6,4,4,6) |

| | | (240,250,256,260,264,270,280) |(10,6,4,4,6,10) |

| | | (236,240,250,256,260,264,270,280,284)|(4,10,6,4,4,4,10,4)|

| 440 | 1 | (436,440,444) |(4, 4) |

...

Former name was:

Numbers j = (i + k)/2 such that i^2+1, j^2+1 and k^2+1 are three consecutive primes.- Robert Israel, Jun 19 2019

Examples

			10 is in the sequence because from the triple (i, j, k) = (6, 10, 14) with j = (i + k)/2 = (6+14)/2 = 10, we obtain the three consecutive primes (i^2+1, j^2+1, k^2+1) = (37, 101, 197).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002496, A005574, A006562.

Programs

Maple

nn:=100000:V:=array(1..6656):kk:=0:
for i from 1 to nn do:
x:=i^2+1:
  if isprime(x)
   then
   kk:=kk+1:V[kk]:=i:
  else
fi:
od:
for n from 2 to kk-2 do:
   p:=V[n]:m:=V[n+1]:q:=V[n+2]:
    if (p+q)/2 = m
     then
      ii:=1:printf(`%d, `,V[n+1]):
    else
   fi:
od:

Mathematica

P = Select[Range[10^4]^2+1, PrimeQ]; Reap[Do[{i, j, k} = Sqrt[P[[n ;; n+2]] - 1]; If[AllTrue[{i, j, k}, IntegerQ] && (i+k)/2 == j, Print[{i, j, k}]; Sow[j]], {n, 1, Length[P]-2}]][[2, 1]] (* Jean-François Alcover, Nov 08 2016 *)

Extensions

Name changed by Robert Israel, Jun 19 2019