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The sum of three consecutive prime numbers, beginning with a(n), is a cube.
3

%I #21 May 22 2020 19:59:34

%S 439,34603,1016201,3696493,4002991,6344687,10221397,14662309,16209029,

%T 19925483,20856907,22805969,43441271,60120691,60761413,62056457,

%U 62710787,87791567,96268243,125977651,166225747,170027449

%N The sum of three consecutive prime numbers, beginning with a(n), is a cube.

%H Chai Wah Wu, <a href="/A210205/b210205.txt">Table of n, a(n) for n = 1..10000</a>

%e prime(85) + prime(86) + prime(87) = 439 + 443 + 449 = 1331 = 11^3.

%t t = {}; p = 2; q = 3; Do[r = NextPrime[q]; If[IntegerQ[(p + q + r)^(1/3)], AppendTo[t, p]; Print[p]]; p = q; q = r, {1000000}]; t (* _T. D. Noe_, Mar 24 2012 *)

%t Select[Partition[Prime[Range[9505000]],3,1],IntegerQ[Surd[Total[#],3]]&][[All,1]] (* _Harvey P. Dale_, May 22 2020 *)

%o (Python)

%o from __future__ import division

%o from sympy import nextprime, prevprime

%o A210205_list = []

%o for i in range(3,10**6):

%o n = i**3

%o p2 = prevprime(n//3)

%o p1, p3 = prevprime(p2), nextprime(p2)

%o q = p1+p2+p3

%o while q <= n:

%o if q == n:

%o A210205_list.append(p1)

%o p1, p2, p3 = p2, p3, nextprime(p3)

%o q = p1+p2+p3 # _Chai Wah Wu_, Dec 31 2015

%Y Cf. A061308.

%K nonn

%O 1,1

%A _Pablo Martínez_, Mar 18 2012

%E Extended by _T. D. Noe_, Mar 24 2012