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A122729
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Primes that are the sum of 5 positive cubes.
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2
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5, 19, 31, 59, 71, 83, 89, 97, 101, 109, 127, 131, 157, 181, 199, 227, 233, 241, 251, 257, 269, 281, 283, 293, 307, 331, 347, 349, 353, 373, 379, 409, 421, 431, 433, 443, 449, 461, 487, 499, 503, 523, 541, 557, 563, 569, 587, 593, 599, 601, 619, 631, 647, 661
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OFFSET
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1,1
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COMMENTS
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By parity, there must be an odd number of odds in the sum. Hence this sequence is the union of primes which are the sum of five odd cubes (such as 5 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3); primes which are the sum of the cube of two even numbers and the cubes of three odd numbers (such as 19 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3); and the primes which are the sum of the cube of an odd number and the cubes of four even numbers (such as 59 = 3^3 + 2^3 + 2^3 + 2^3 + 2^3). A subset of this sequence is the primes which are the sum of the cubes of five distinct primes (i.e. of the form p^3 + q^3 + r^3 + s^3 + t^3 for p, q, r, s, t distinct odd primes) such as 105649 = 3^3 + 5^3 + 7^3 + 11^3 + 47^3. No prime can be the sum of two cubes (by factorization of the sum of two cubes).
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LINKS
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FORMULA
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EXAMPLE
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a(1) = 5 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3.
a(2) = 19 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3.
a(3) = 31 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3.
a(4) = 59 = 3^3 + 2^3 + 2^3 + 2^3 + 2^3.
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MATHEMATICA
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q = 10; lst = {}; Do[Do[Do[Do[Do[p = a^3 + b^3 + c^3 + d^3 + e^3; If[PrimeQ[p], AppendTo[lst, p]], {e, q}], {d, q}], {c, q}], {b, q}], {a, q}]; Take[Union[lst], 80] (* Vladimir Joseph Stephan Orlovsky, Jul 15 2011 *)
With[{upto=650}, Union[Select[Total/@Tuples[Range[Surd[upto-4, 3]]^3, 5], PrimeQ[ #]&&#<=upto&]]] (* Harvey P. Dale, Sep 30 2018 *)
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PROG
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(PARI) list(lim)=my(ta, tb, tc, td, te, v=List()); for(a=1, (lim/5)^(1/3), ta=a^3; for(b=a, ((lim-ta)/4)^(1/3), tb=ta+b^3; for(c=b, ((lim-tb)/3)^(1/3), tc=tb+c^3; for(d=c, ((lim-tc)/2)^(1/3), td=tc+d^3; forstep(e=if(td%2==d%2, d+1, d), (lim-td)^(1/3), 2, te=td+e^3; if(ispseudoprime(te), listput(v, te))))))); vecsort(Vec(v), , 8) \\ Charles R Greathouse IV, Jul 15 2011
(Python)
from sympy import isprime
from collections import Counter
from itertools import combinations_with_replacement as combs_w_rep
def aupto(lim):
s = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))
s2 = filter(lambda x: x<=lim, (sum(c) for c in combs_w_rep(s, 5)))
s2counts = Counter(s2)
return sorted(filter(isprime, s2counts))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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