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A248587
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The smallest of four consecutive primes whose sum is a perfect cube.
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1
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4812191, 6353029, 8039333, 8821867, 19876711, 60742631, 85017061, 108879847, 127042367, 138853049, 170367959, 238190951, 259108427, 414949357, 485941193, 512095739, 529218559, 582868471, 623331491, 648485381, 771656657, 1001132351, 1098706507, 1172752457
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OFFSET
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1,1
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LINKS
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EXAMPLE
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a(2) = 6353029 is prime. Next three primes are 6353033, 6353051 and 6353071. Their sum = 6353029 + 6353033 + 6353051 + 6353071 = 25412184 = 294^3.
a(3) = 8039333 is prime. Next three primes are 8039359, 8039363 and 8039377. Their sum = 8039333 + 8039359 + 8039363 + 8039377 = 32157432 = 318^3.
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MATHEMATICA
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t = {}; p = 2; q = 3; r = 5; Do[v = NextPrime[r]; If[IntegerQ[(p + q + r + v)^(1/3)], AppendTo[t, p]; Print[p]]; p = q; q = r; r = v, {5*10^8}]; t
Select[Partition[Prime[Range[6*10^7]], 4, 1], IntegerQ[Surd[Total[#], 3]] &] [[All, 1]] (* Harvey P. Dale, Oct 07 2016 *)
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PROG
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(PARI) lista(nn) = {vp = primes(nn); for (i=1, #vp - 3, if (ispower(vp[i]+vp[i+1]+vp[i+2]+vp[i+3], 3), print1(vp[i], ", ")); ); } \\ Michel Marcus, Oct 24 2014
(Python)
from sympy import nextprime, prevprime
for i in range(3, 10**6):
n = i**3
p3 = prevprime(n//4)
p2, p4 = prevprime(p3), nextprime(p3)
p1 = prevprime(p2)
q = p1+p2+p3+p4
while q <= n:
if q == n:
p1, p2, p3, p4 = p2, p3, p4, nextprime(p4)
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CROSSREFS
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Cf. A061308 (two consecutive primes), A210205 (three consecutive primes).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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