A292715 Initial member of 8 consecutive primes a, b, c, d, e, f, g, h such that (a + h) = (b + g), (c + g) = (d + f), (a + f) = (b + e) and (a + g) = (b + f).

6337, 14717, 77521, 83401, 130643, 344231, 357653, 380377, 496453, 505067, 587101, 593473, 970457, 1130251, 1515691, 1694191, 1936741, 2689997, 2773007, 2811163, 3665371, 3678887, 3713993, 3976361, 4024687, 4181579, 4629461, 4801673, 5438569, 5882197, 6016811

Offset: 1

Author

K. D. Bajpai, Sep 21 2017

Keywords

• nonn

Comments

The equations reduce to b-a = d-c = f-e = g-f = h-g, i.e. gap number 1, 3, 5, 6, 7 are equal. - Jens Kruse Andersen, Oct 14 2017

Examples

			6337 is a term because it is the first of 8 consecutive primes {6337, 6343, 6353, 6359, 6361, 6367, 6373, 6379} = {a, b, c, d, e, f, g, h}; and (a + h) = (b + g), (c + g) = (d + f), (a + f) = (b + e) and (a + g) = (b + f).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..230

Crossrefs

Cf. A000040, A292618.

Programs

Maple

A292715:= proc(n)local a, b, c, d, e, f, g, h; a:=ithprime(n); b:=ithprime(n + 1); c:=ithprime(n + 2); d:=ithprime(n + 3); e:=ithprime(n + 4); f:=ithprime(n + 5); g:=ithprime(n + 6); h:=ithprime(n + 7); if (a + h) = (b + g) and (c + g) = (d + f) and (a + f) = (b + e) and (a + g)=(b + f) then RETURN (a); fi; end: seq(A292715(n), n=1..1000000);

Mathematica

Select[Partition[Prime@ Range[10^6], 8, 1], Function[{a, b, c, d, e, f, g, h}, And[(a + h) == (b + g), (c + g) == (d + f), (a + f) == (b + e), (a + g) == (b + f)]] @@ # &][[All, 1]] (* Michael De Vlieger, Sep 21 2017 *)
Select[Partition[Prime[Range[420000]],8,1],Length[Union[Drop[Drop[ Differences[ #],{4}],{2}]]]==1&][[All,1]](* Harvey P. Dale, Nov 03 2017 *)