A345599 Numbers that are the sum of ten fourth powers in six or more ways.

3175, 4150, 4230, 4390, 4405, 4455, 4470, 4485, 4500, 4550, 4565, 4630, 4725, 4740, 4915, 4980, 5094, 5109, 5155, 5190, 5205, 5220, 5270, 5285, 5350, 5365, 5395, 5430, 5445, 5460, 5475, 5525, 5540, 5590, 5605, 5635, 5655, 5670, 5700, 5715, 5735, 5765, 5780

Offset

1,1

Links

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

Example

4150 is a term because 4150 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 = 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 7^4.

Prog

(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1, 1000)]
for pos in cwr(power_terms, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 6])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345554, A345590, A345598, A345600, A345638, A345858.

Sequence in context: A217247 A020420 A155484 * A345858 A129472 A023309

Adjacent sequences: A345596 A345597 A345598 * A345600 A345601 A345602

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved