A345566 Numbers that are the sum of six fourth powers in nine or more ways.
88595, 122915, 132546, 134931, 144835, 146450, 151556, 161475, 162355, 162755, 170275, 171555, 171795, 172036, 172835, 173075, 177380, 177716, 180770, 183540, 183620, 184835, 185315, 185555, 187700, 187715, 190100, 190211, 193635, 195380, 195780, 196435
Offset: 1
Keywords
Examples
122915 is a term because 122915 = 1^4 + 3^4 + 6^4 + 9^4 + 10^4 + 18^4 = 1^4 + 4^4 + 7^4 + 8^4 + 15^4 + 16^4 = 1^4 + 7^4 + 9^4 + 10^4 + 14^4 + 16^4 = 2^4 + 3^4 + 4^4 + 5^4 + 14^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 11^4 + 18^4 = 2^4 + 9^4 + 9^4 + 12^4 + 14^4 + 15^4 = 3^4 + 5^4 + 6^4 + 6^4 + 11^4 + 18^4 = 3^4 + 8^4 + 10^4 + 11^4 + 13^4 + 16^4 = 5^4 + 6^4 + 7^4 + 11^4 + 14^4 + 16^4 = 8^4 + 8^4 + 9^4 + 10^4 + 11^4 + 17^4.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..10000
Programs
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Python
from itertools import combinations_with_replacement as cwr from collections import defaultdict keep = defaultdict(lambda: 0) power_terms = [x**4 for x in range(1, 1000)] for pos in cwr(power_terms, 6): tot = sum(pos) keep[tot] += 1 rets = sorted([k for k, v in keep.items() if v >= 9]) for x in range(len(rets)): print(rets[x])