A346333 Numbers that are the sum of eight fifth powers in exactly eight ways.
8625619, 9773236, 10036233, 10071050, 12247994, 13180706, 13377868, 13662501, 14584992, 14591744, 14611077, 15251119, 16112362, 16374250, 16391025, 16472544, 16588000, 16667851, 17059075, 17216298, 17405300, 17917097, 18107564, 18392902, 18470839, 18541635
Offset: 1
Keywords
Examples
8625619 is a term because 8625619 = 2^5 + 5^5 + 5^5 + 9^5 + 10^5 + 12^5 + 12^5 + 24^5 = 1^5 + 3^5 + 8^5 + 9^5 + 11^5 + 11^5 + 12^5 + 24^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 16^5 + 23^5 = 1^5 + 3^5 + 3^5 + 4^5 + 11^5 + 17^5 + 18^5 + 22^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 16^5 + 22^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 16^5 + 19^5 + 20^5 = 3^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 18^5 + 20^5 = 3^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 18^5 + 20^5.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..10000
Programs
-
Python
from itertools import combinations_with_replacement as cwr from collections import defaultdict keep = defaultdict(lambda: 0) power_terms = [x**5 for x in range(1, 1000)] for pos in cwr(power_terms, 8): 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])
Comments