A344035 Numbers that are the sum of five positive cubes in exactly four ways.

1252, 1376, 1461, 1522, 1548, 1585, 1590, 1646, 1702, 1709, 1737, 1739, 1772, 1798, 1802, 1810, 1864, 1889, 1954, 1987, 2006, 2033, 2081, 2096, 2152, 2160, 2225, 2241, 2251, 2276, 2313, 2322, 2339, 2341, 2367, 2374, 2377, 2416, 2423, 2456, 2458, 2465, 2467, 2512, 2521, 2528, 2530, 2537, 2540, 2549, 2556, 2582

Offset: 1

Author

David Consiglio, Jr., May 07 2021

Keywords

• nonn

Comments

Differs from A344034 at term 13 because 1765 = 1^3 + 1^3 + 2^3 + 3^3 + 12^3 = 1^3 + 1^3 + 6^3 + 6^3 + 11^3 = 1^3 + 2^3 + 3^3 + 9^3 + 10^3 = 3^3 + 4^3 + 6^3 + 9^3 + 9^3 = 4^3 + 4^3 + 5^3 + 8^3 + 10^3

Examples

			1461 is a member of this sequence because 1461 = 1^3 + 1^3 + 1^3 + 9^3 + 9^3 = 1^3 + 1^3 + 4^3 + 4^3 + 11^3 = 3^3 + 3^3 + 4^3 + 7^3 + 10^3 = 6^3 + 6^3 + 7^3 + 7^3 + 7^3
		

Links

• David Consiglio, Jr., Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A294738, A343705, A343972, A343988, A344034, A344355, A345766.

Programs

Mathematica

s5pcQ[n_]:=Length[Select[PowersRepresentations[n,5,3],FreeQ[#,0]&]]==4; Select[Range[ 3000],s5pcQ] (* Harvey P. Dale, Sep 15 2024 *)

Python

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1,50)]
for pos in cwr(power_terms,5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k,v in keep.items() if v == 4])
for x in range(len(rets)):
    print(rets[x])