A345626 Numbers that are the sum of nine fifth powers in nine or more ways.

1969221, 2596936, 3353186, 3378178, 3923426, 3981447, 4094027, 4096729, 4112329, 4114188, 4129465, 4137209, 4147736, 4157156, 4170112, 4172994, 4254304, 4303773, 4410482, 4475846, 4477936, 4483379, 4485480, 4492410, 4501441, 4510461, 4543232, 4652011, 4691855

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			2596936 is a term because 2596936 = 1^5 + 1^5 + 4^5 + 5^5 + 9^5 + 13^5 + 13^5 + 13^5 + 17^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 17^5 = 1^5 + 4^5 + 7^5 + 7^5 + 7^5 + 9^5 + 9^5 + 14^5 + 18^5 = 1^5 + 5^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 12^5 + 13^5 + 14^5 + 17^5 = 2^5 + 5^5 + 5^5 + 6^5 + 6^5 + 6^5 + 15^5 + 15^5 + 16^5 = 3^5 + 3^5 + 5^5 + 6^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 11^5 + 11^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 8^5 + 9^5 + 16^5 + 17^5.
		

Links

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

Crossrefs

Cf. A345593, A345617, A345625, A345627, A345641, A346344.

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 >= 9])
    for x in range(len(rets)):
        print(rets[x])