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

1,1

Links

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

Example

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.

Prog

(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])

Crossrefs

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

Sequence in context: A254559 A254482 A219434 * A346344 A172791 A233996

Adjacent sequences: A345623 A345624 A345625 * A345627 A345628 A345629

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved