A345630 Numbers that are the sum of seven fifth powers in eight or more ways.

36620574, 80552143, 81401376, 82078424, 92347417, 93653176, 94486699, 94626949, 98873875, 105674625, 110276376, 121050874, 124732805, 125959393, 127808693, 129228307, 130298618, 134581976, 144209018, 145340799, 147245218, 147898763, 151727082

Offset: 1

Author

David Consiglio, Jr., Jun 22 2021

Keywords

• nonn

Examples

			80552143 is a term because 80552143 = 1^5 + 4^5 + 21^5 + 21^5 + 23^5 + 29^5 + 34^5 = 1^5 + 8^5 + 14^5 + 23^5 + 23^5 + 32^5 + 32^5 = 1^5 + 8^5 + 16^5 + 19^5 + 27^5 + 28^5 + 34^5 = 3^5 + 12^5 + 13^5 + 14^5 + 28^5 + 31^5 + 32^5 = 3^5 + 14^5 + 17^5 + 18^5 + 18^5 + 27^5 + 36^5 = 4^5 + 11^5 + 13^5 + 22^5 + 23^5 + 24^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 23^5 + 24^5 + 36^5 = 6^5 + 23^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5.
		

Links

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

Crossrefs

Cf. A345574, A345616, A345629, A345631, A345722, A346285.

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])