A346344 Numbers that are the sum of nine fifth powers in exactly nine ways.
1969221, 2596936, 3353186, 3378178, 3923426, 3981447, 4094027, 4096729, 4112329, 4114188, 4129465, 4137209, 4147736, 4170112, 4172994, 4254304, 4303773, 4410482, 4475846, 4477936, 4483379, 4485480, 4501441, 4543232, 4652011, 4691855, 4724015, 4733970, 4750241
Offset: 1
Keywords
Examples
1969221 is a term because 1969221 = 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 + 16^5 = 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 12^5 + 12^5 + 13^5 + 16^5 = 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 12^5 + 12^5 + 13^5 + 16^5 = 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 14^5 + 15^5 = 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 14^5 + 15^5 = 1^5 + 4^5 + 5^5 + 8^5 + 9^5 + 13^5 + 13^5 + 13^5 + 15^5 = 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 + 14^5 = 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 + 14^5.
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**5 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])
Comments