A345623 Numbers that are the sum of nine fifth powers in six or more ways.

926404, 936607, 952896, 985421, 993574, 993605, 993816, 1075779, 1123321, 1133344, 1134367, 1151406, 1160105, 1166111, 1177144, 1206514, 1209669, 1209847, 1215545, 1225630, 1251130, 1264929, 1265320, 1278611, 1414834, 1422367, 1422609, 1430384, 1431367

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			936607 is a term because 936607 = 1^5 + 1^5 + 2^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 12^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 9^5 + 11^5 + 11^5 + 14^5 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 8^5 + 10^5 + 12^5 + 14^5 = 4^5 + 4^5 + 4^5 + 7^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A345590, A345614, A345622, A345624, A345638, A346341.

Programs

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])