A345149 - OEIS

A345149

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

6883, 12411, 13923, 14112, 14581, 14896, 14904, 15561, 15876, 16317, 16640, 17208, 17479, 17992, 18739, 18865, 19035, 19080, 19665, 19712, 19763, 19880, 20007, 20384, 20979, 21231, 21420, 21707, 22409, 22617, 23149, 23940, 24355, 25515, 25984, 26208, 26334

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OFFSET

1,1

COMMENTS

Differs from A345148 at term 3 because 13104 = 1^3 + 10^3 + 16^3 + 18^3 = 1^3 + 11^3 + 14^3 + 19^3 = 2^3 + 9^3 + 15^3 + 19^3 = 4^3 + 6^3 + 14^3 + 20^3 = 4^3 + 9^3 + 10^3 + 21^3 = 5^3 + 7^3 + 11^3 + 21^3 = 8^3 + 9^3 + 14^3 + 19^3.

LINKS

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

EXAMPLE

6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3 = 2^3 + 4^3 + 14^3 + 14^3 = 3^3 + 7^3 + 7^3 + 17^3 = 3^3 + 10^3 + 13^3 + 13^3 = 4^3 + 10^3 + 10^3 + 15^3 = 7^3 + 8^3 + 8^3 + 16^3.

PROG

(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 == 6])

for x in range(len(rets)):

print(rets[x])