A345614 Numbers that are the sum of eight fifth powers in six or more ways.

1431397, 2593811, 3329119, 3345410, 3609912, 3800722, 3932480, 4093604, 4096697, 4104553, 4114187, 4129433, 4154031, 4169869, 4377714, 4451412, 4475603, 4484634, 4501409, 4730845, 4756642, 4882770, 4912477, 4915506, 4970823, 5003645, 5112274, 5259111, 5449985

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			2593811 is a term because 2593811 = 1^5 + 1^5 + 4^5 + 9^5 + 13^5 + 13^5 + 13^5 + 17^5 = 1^5 + 1^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 17^5 = 1^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 = 2^5 + 5^5 + 6^5 + 6^5 + 6^5 + 15^5 + 15^5 + 16^5 = 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 6^5 + 11^5 + 11^5 + 13^5 + 18^5.
		

Links

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

Crossrefs

Cf. A345581, A345609, A345613, A345615, A345623, A346331.

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, 8):
    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])