[go: up one dir, main page]

login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A345149
Numbers that are the sum of four third powers in exactly six ways.
7
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
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])
KEYWORD
nonn
AUTHOR
STATUS
approved