A345629 Numbers that are the sum of seven fifth powers in seven or more ways.

28608832, 35663099, 36090526, 36620574, 46998599, 51095638, 52541851, 54233651, 54827543, 54886349, 61263643, 61634374, 63514593, 64810976, 65198607, 66708676, 67887843, 70979107, 72970305, 74002457, 74115801, 74132607, 74487093, 75044651, 75378359

Offset: 1

Author

David Consiglio, Jr., Jun 22 2021

Keywords

• nonn

Examples

			35663099 is a term because 35663099 = 1^5 + 9^5 + 16^5 + 17^5 + 24^5 + 24^5 + 28^5 = 2^5 + 3^5 + 17^5 + 23^5 + 24^5 + 24^5 + 26^5 = 2^5 + 10^5 + 15^5 + 17^5 + 23^5 + 23^5 + 29^5 = 4^5 + 8^5 + 13^5 + 19^5 + 21^5 + 27^5 + 27^5 = 4^5 + 11^5 + 13^5 + 19^5 + 20^5 + 22^5 + 30^5 = 5^5 + 6^5 + 19^5 + 19^5 + 20^5 + 20^5 + 30^5 = 7^5 + 9^5 + 12^5 + 18^5 + 18^5 + 27^5 + 28^5.
		

Links

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

Crossrefs

Cf. A345573, A345609, A345615, A345630, A345721, A346284.

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, 7):
    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])