A345641 Numbers that are the sum of ten fifth powers in nine or more ways.
1192180, 1226654, 1242437, 1431399, 1431430, 1431641, 1431672, 1431883, 1432453, 1432664, 1434765, 1439174, 1439416, 1441695, 1442718, 1447602, 1448447, 1455346, 1455377, 1464166, 1464377, 1464408, 1474431, 1474462, 1475485, 1491978, 1497619, 1505660, 1531398
Offset: 1
Keywords
Examples
1226654 is a term because 1226654 = 1^5 + 1^5 + 4^5 + 5^5 + 7^5 + 8^5 + 9^5 + 13^5 + 13^5 + 13^5 = 1^5 + 1^5 + 5^5 + 7^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 = 1^5 + 2^5 + 2^5 + 4^5 + 6^5 + 10^5 + 12^5 + 12^5 + 12^5 + 13^5 = 1^5 + 2^5 + 2^5 + 4^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 14^5 = 1^5 + 3^5 + 3^5 + 3^5 + 7^5 + 10^5 + 10^5 + 10^5 + 13^5 + 14^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 7^5 + 8^5 + 12^5 + 13^5 + 14^5 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 5^5 + 10^5 + 13^5 + 13^5 + 13^5 = 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5 + 8^5 + 9^5 + 16^5 = 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 8^5 + 9^5 + 16^5.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..1000
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 >= 9]) for x in range(len(rets)): print(rets[x])