A122729 Primes that are the sum of 5 positive cubes.

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

Offset: 1

Author

Jonathan Vos Post, Sep 23 2006

Keywords

• easy
• nonn

Comments

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).

Examples

			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.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A003328.

Programs

Mathematica

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 *)

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))
print(aupto(661)) # Michael S. Branicky, May 19 2021

Formula

A000040 INTERSECTION A003328.