A346353 Numbers that are the sum of ten fifth powers in exactly eight ways.
944383, 953139, 953414, 985453, 1118585, 1151438, 1185375, 1198879, 1206546, 1209912, 1216569, 1217172, 1218912, 1223321, 1225398, 1234631, 1241834, 1251195, 1251406, 1252123, 1259685, 1265563, 1265594, 1267937, 1275375, 1281736, 1295418, 1297697, 1298088
Offset: 1
Keywords
Examples
944383 is a term because 944383 = 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 + 10^5 + 12^5 + 14^5 = 2^5 + 4^5 + 5^5 + 5^5 + 7^5 + 7^5 + 7^5 + 10^5 + 12^5 + 14^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 12^5.
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 == 8]) for x in range(len(rets)): print(rets[x])
Comments