A345156 Numbers that are the sum of four third powers in exactly ten ways.

21896, 36225, 48825, 51506, 52416, 53200, 58338, 58968, 60480, 66024, 67851, 70434, 70525, 71155, 72819, 76923, 78624, 78912, 85995, 87507, 88641, 90181, 90783, 91728, 93555, 97552, 98280, 98560, 99008, 99225, 99792, 100170, 103040, 104104, 104265, 104958

Offset: 1

Author

David Consiglio, Jr., Jun 09 2021

Keywords

• nonn

Comments

Differs from A345155 at term 3 because 46872 = 1^3 + 16^3 + 22^3 + 30^3 = 2^3 + 11^3 + 17^3 + 33^3 = 3^3 + 3^3 + 4^3 + 35^3 = 3^3 + 4^3 + 26^3 + 29^3 = 3^3 + 5^3 + 23^3 + 31^3 = 4^3 + 10^3 + 24^3 + 30^3 = 5^3 + 17^3 + 23^3 + 29^3 = 6^3 + 10^3 + 20^3 + 32^3 = 11^3 + 11^3 + 21^3 + 31^3 = 11^3 + 14^3 + 17^3 + 32^3 = 19^3 + 21^3 + 21^3 + 25^3.

Examples

			21896 is a term because 21896 = 1^3 + 11^3 + 19^3 + 22^3  = 2^3 + 2^3 + 12^3 + 26^3  = 2^3 + 3^3 + 19^3 + 23^3  = 2^3 + 5^3 + 15^3 + 25^3  = 3^3 + 10^3 + 16^3 + 24^3  = 3^3 + 17^3 + 19^3 + 19^3  = 4^3 + 6^3 + 20^3 + 22^3  = 5^3 + 8^3 + 14^3 + 25^3  = 7^3 + 11^3 + 17^3 + 23^3  = 8^3 + 9^3 + 19^3 + 22^3.
		

Links

• David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A025366, A344929, A345122, A345154, A345155, A345188.

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, 4):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 10])
for x in range(len(rets)):
    print(rets[x])