A345851 Numbers that are the sum of nine fourth powers in exactly nine ways.
8259, 9539, 10709, 10819, 10884, 10949, 10964, 11124, 11444, 11573, 11668, 11684, 11924, 12099, 12164, 12339, 12404, 12549, 12773, 12853, 12918, 13013, 13139, 13204, 13284, 13379, 13444, 13509, 13958, 13988, 14053, 14213, 14293, 14308, 14373, 14403, 14484
Offset: 1
Keywords
Examples
9299 is a term because 9299 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 7^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 9^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4.
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, 9): 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])
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