A345088 Numbers that are the sum of three third powers in exactly eight ways.

2562624, 5618250, 6525000, 6755328, 7374375, 12742920, 13581352, 14027112, 14288373, 18443160, 20500992, 22783032, 23113728, 25305048, 26936064, 27131840, 29515968, 30205440, 32835375, 38269440, 39317832, 39339000, 40189248, 42144192, 42183504, 43077952

Offset: 1

Author

David Consiglio, Jr., Jun 07 2021

Keywords

• nonn

Comments

Differs from A345087 at term 10 because 14926248 = 2^3 + 33^3 + 245^3 = 11^3 + 185^3 + 203^3 = 14^3 + 32^3 + 245^3 = 50^3 + 113^3 + 236^3 = 71^3 + 89^3 + 239^3 = 74^3 + 189^3 + 196^3 = 89^3 + 185^3 + 197^3 = 98^3 + 148^3 + 219^3 = 105^3 + 149^3 + 217^3.

Examples

			2562624 is a term because 2562624 = 7^3 + 35^3 + 135^3  = 7^3 + 63^3 + 131^3  = 11^3 + 99^3 + 115^3  = 16^3 + 45^3 + 134^3  = 29^3 + 102^3 + 112^3  = 35^3 + 59^3 + 131^3  = 50^3 + 84^3 + 121^3  = 68^3 + 71^3 + 122^3.
		

Links

• David Consiglio, Jr., Table of n, a(n) for n = 1..100

Crossrefs

Cf. A025328, A344738, A345085, A345087, A345120, A345153.

Programs

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, 1000)]
for pos in cwr(power_terms, 3):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 8])
for x in range(len(rets)):
    print(rets[x])