A346917 Numbers that are a sum of the cubes of four primes, not necessarily distinct.

32, 51, 70, 89, 108, 149, 168, 187, 206, 266, 285, 304, 367, 383, 386, 402, 405, 424, 484, 500, 503, 522, 601, 620, 702, 718, 721, 740, 819, 838, 936, 1037, 1056, 1154, 1355, 1372, 1374, 1393, 1412, 1472, 1491, 1510, 1589, 1608, 1690, 1706, 1709, 1728, 1807, 1826

Offset: 1

Author

Amiram Eldar, Aug 07 2021

Keywords

• nonn

Comments

Roth (1951) proved that the number of terms below x is >> x/log(x)^8.

Ren (2001) proved that this sequence has a positive lower density.

The lower density was proven to be larger than 0.003125 (Ren, 2003), 0.005776 (Liu, 2012), and 0.009664 (Elsholtz and Schlage-Puchta, 2019).

Examples

			a(1) = 32 = 2^3 + 2^3 + 2^3 + 2^3.
a(2) = 51 = 2^3 + 2^3 + 2^3 + 3^3.
a(3) = 70 = 2^3 + 2^3 + 3^3 + 3^3.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000
• Christian Elsholtz and Jan-Christoph Schlage-Puchta, The density of integers representable as the sum of four prime cubes, arXiv preprint, arXiv:1902.09858 [math.NT], 2019.
• Zhixin Liu, Density of the sums of four cubes of primes, Journal of Number Theory, Vol. 132, No. 4 (2012), pp. 735-747.
• Xiumin Ren, Density of integers that are the sum of four cubes of primes, Chin. Ann. Math. Ser. B, Vol. 22, No. 2 (2001), pp. 233-242.
• Xiumin Ren, Sums of four cubes of primes, J. Number Theory, Vol. 98, No. 1 (2003), pp. 156-171.
• K. F. Roth, On Waring's problem for cubes, Proc. London Math. Soc. (2), Vol. 53 (1951), pp. 268-279.

Crossrefs

Cf. A030078, A086119, A258865.

Programs

Mathematica

seq[max_] := Module[{s = Select[Range[Floor @ Surd[max, 3]], PrimeQ]}, Select[Union[Plus @@@ (Tuples[s, 4]^3)], # <= max &]]; seq[2000]

PARI

list(lim)=my(v=List(), P=apply(p->p^3,primes(sqrtnint(lim\=1,3)))); foreach(P,p, foreach(P,q, foreach(P,r, my(s=p+q+r,t); for(i=1,#P, t=s+P[i]; if(t>lim, break); listput(v,t))))); Set(v) \\ Charles R Greathouse IV, Aug 09 2021

Python

from sympy import integer_nthroot, primerange
from itertools import combinations_with_replacement as cwr
def aupto(limit):
    cubes = [p**3 for p in primerange(2, integer_nthroot(limit, 3)[0])]
    return sorted(sum(c) for c in cwr(cubes, 4) if sum(c) <= limit)
print(aupto(2000)) # Michael S. Branicky, Apr 09 2022