login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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
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
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-François Alcover, Nov 08 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Nov 07 2016
EXTENSIONS
Name changed by Robert Israel, Jun 19 2019
STATUS
approved