This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A277970 #19 Jun 20 2019 02:53:33
%S A277970 4,10,20,120,130,180,230,260,440,470,680,700,750,920,1060,1320,1736,
%T A277970 1860,1970,2106,2320,2460,2760,2850,2890,3074,3660,3800,4180,4370,
%U A277970 5030,5236,5304,5814,5990,6130,6350,6590,6724,6780,6990,7190,7384,7520,7744,8180
%N A277970 Numbers k = A005574(m) such that k = (A005574(m-1)+A005574(m+1))/2.
%C A277970 Subsequence of A005574.
%C A277970 a(n) == 0, 4 or 6 (mod 10).
%C A277970 The corresponding primes a(n)^2+1 are 17, 101, 401, 14401, 16901, ...
%C A277970 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.
%C A277970 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.
%C A277970 +------+-----+--------------------------------------+-------------------+
%C A277970 | a(n) |order| m-tuples | differences |
%C A277970 +------+-----+--------------------------------------+-------------------+
%C A277970 | 4 | 1 | (2,4,6) |(2, 2) |
%C A277970 | 10 | 2 | (6,10,14) |(4, 4) |
%C A277970 | | | (4,6,10,14,16) |(2,4,4,2) |
%C A277970 | 20 | 1 | (16,20,24) |(4,4) |
%C A277970 | 120 | 1 | (116,120,124) |(4,4) |
%C A277970 | 130 | 1 | (126,130,134) |(4,4) |
%C A277970 | 180 | 1 | (176,180,184) |(4,4) |
%C A277970 | 230 | 1 | (224,230,236) |(6,6) |
%C A277970 | 260 | 4 | (256,260,264) |(4,4) |
%C A277970 | | | (250,256,260,264,270) |(6,4,4,6) |
%C A277970 | | | (240,250,256,260,264,270,280) |(10,6,4,4,6,10) |
%C A277970 | | | (236,240,250,256,260,264,270,280,284)|(4,10,6,4,4,4,10,4)|
%C A277970 | 440 | 1 | (436,440,444) |(4, 4) |
%C A277970 ...
%C A277970 Former name was:
%C A277970 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
%H A277970 Robert Israel, <a href="/A277970/b277970.txt">Table of n, a(n) for n = 1..10000</a>
%e A277970 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).
%p A277970 nn:=100000:V:=array(1..6656):kk:=0:
%p A277970 for i from 1 to nn do:
%p A277970 x:=i^2+1:
%p A277970 if isprime(x)
%p A277970 then
%p A277970 kk:=kk+1:V[kk]:=i:
%p A277970 else
%p A277970 fi:
%p A277970 od:
%p A277970 for n from 2 to kk-2 do:
%p A277970 p:=V[n]:m:=V[n+1]:q:=V[n+2]:
%p A277970 if (p+q)/2 = m
%p A277970 then
%p A277970 ii:=1:printf(`%d, `,V[n+1]):
%p A277970 else
%p A277970 fi:
%p A277970 od:
%t A277970 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 *)
%Y A277970 Cf. A002496, A005574, A006562.
%K A277970 nonn
%O A277970 1,1
%A A277970 _Michel Lagneau_, Nov 07 2016
%E A277970 Name changed by _Robert Israel_, Jun 19 2019