|
%I #19 Jun 20 2019 02:53:33
%S 4,10,20,120,130,180,230,260,440,470,680,700,750,920,1060,1320,1736,
%T 1860,1970,2106,2320,2460,2760,2850,2890,3074,3660,3800,4180,4370,
%U 5030,5236,5304,5814,5990,6130,6350,6590,6724,6780,6990,7190,7384,7520,7744,8180
%N Numbers k = A005574(m) such that k = (A005574(m-1)+A005574(m+1))/2.
%C Subsequence of A005574.
%C a(n) == 0, 4 or 6 (mod 10).
%C The corresponding primes a(n)^2+1 are 17, 101, 401, 14401, 16901, ...
%C 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 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 +------+-----+--------------------------------------+-------------------+
%C | a(n) |order| m-tuples | differences |
%C +------+-----+--------------------------------------+-------------------+
%C | 4 | 1 | (2,4,6) |(2, 2) |
%C | 10 | 2 | (6,10,14) |(4, 4) |
%C | | | (4,6,10,14,16) |(2,4,4,2) |
%C | 20 | 1 | (16,20,24) |(4,4) |
%C | 120 | 1 | (116,120,124) |(4,4) |
%C | 130 | 1 | (126,130,134) |(4,4) |
%C | 180 | 1 | (176,180,184) |(4,4) |
%C | 230 | 1 | (224,230,236) |(6,6) |
%C | 260 | 4 | (256,260,264) |(4,4) |
%C | | | (250,256,260,264,270) |(6,4,4,6) |
%C | | | (240,250,256,260,264,270,280) |(10,6,4,4,6,10) |
%C | | | (236,240,250,256,260,264,270,280,284)|(4,10,6,4,4,4,10,4)|
%C | 440 | 1 | (436,440,444) |(4, 4) |
%C ...
%C Former name was:
%C 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 Robert Israel, <a href="/A277970/b277970.txt">Table of n, a(n) for n = 1..10000</a>
%e 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 nn:=100000:V:=array(1..6656):kk:=0:
%p for i from 1 to nn do:
%p x:=i^2+1:
%p if isprime(x)
%p then
%p kk:=kk+1:V[kk]:=i:
%p else
%p fi:
%p od:
%p for n from 2 to kk-2 do:
%p p:=V[n]:m:=V[n+1]:q:=V[n+2]:
%p if (p+q)/2 = m
%p then
%p ii:=1:printf(`%d, `,V[n+1]):
%p else
%p fi:
%p od:
%t 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 Cf. A002496, A005574, A006562.
%K nonn
%O 1,1
%A _Michel Lagneau_, Nov 07 2016
%E Name changed by _Robert Israel_, Jun 19 2019
|
