A346352 Numbers that are the sum of ten fifth powers in exactly seven ways.
555098, 674040, 683166, 707315, 763631, 777852, 778844, 780945, 783224, 893654, 896500, 897668, 920887, 926616, 927819, 928802, 936850, 937631, 945017, 952897, 953077, 953350, 955178, 963131, 975133, 979482, 984133, 985664, 987257, 991908, 993575, 993606
Offset: 1
Keywords
Examples
555098 is a term because 555098 = 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 2^5 + 3^5 + 7^5 + 14^5 = 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 7^5 + 8^5 + 10^5 + 13^5 = 1^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 7^5 + 10^5 + 13^5 = 1^5 + 2^5 + 5^5 + 7^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 13^5 = 4^5 + 4^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 11^5 + 12^5 = 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 7^5 + 9^5 + 9^5 + 11^5 + 12^5 = 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 + 11^5.
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, 10): tot = sum(pos) keep[tot] += 1 rets = sorted([k for k, v in keep.items() if v == 7]) for x in range(len(rets)): print(rets[x])
Comments