A345625 Numbers that are the sum of nine fifth powers in eight or more ways.

1431398, 1431640, 1531397, 1952415, 1969221, 2247917, 2530399, 2596936, 2652563, 2652860, 2736790, 2851254, 2965588, 3088909, 3148674, 3273590, 3297416, 3329120, 3329362, 3332244, 3336895, 3345442, 3345653, 3353186, 3361614, 3362217, 3364738, 3378178, 3553641

Offset

1,1

Links

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

Example

1431640 is a term because 1431640 = 1^5 + 2^5 + 3^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 3^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 1^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 9^5 + 12^5 + 16^5 = 2^5 + 2^5 + 3^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 2^5 + 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 3^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5.

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

Crossrefs

Cf. A345592, A345616, A345624, A345626, A345640, A346343.

Sequence in context: A345614 A346331 A345624 * A346343 A346342 A345642

Adjacent sequences: A345622 A345623 A345624 * A345626 A345627 A345628

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved