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A277970 Numbers k = A005574(m) such that k = (A005574(m-1)+A005574(m+1))/2. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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

LINKS

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

EXAMPLE

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).

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-Fran├žois Alcover, Nov 08 2016 *)

CROSSREFS

Cf. A002496, A005574, A006562.

Sequence in context: A038065 A249975 A306158 * A229884 A038422 A009870

Adjacent sequences:  A277967 A277968 A277969 * A277971 A277972 A277973

KEYWORD

nonn

AUTHOR

Michel Lagneau, Nov 07 2016

EXTENSIONS

Name changed by Robert Israel, Jun 19 2019

STATUS

approved

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Last modified July 18 15:43 EDT 2019. Contains 325144 sequences. (Running on oeis4.)