A345601 Numbers that are the sum of ten fourth powers in eight or more ways.

6675, 6740, 6755, 6805, 6820, 6870, 6885, 6950, 6995, 7015, 7030, 7045, 7060, 7095, 7110, 7125, 7270, 7285, 7300, 7350, 7365, 7429, 7494, 7525, 7540, 7590, 7605, 7750, 7780, 7845, 7860, 7925, 7955, 7990, 8005, 8020, 8035, 8085, 8100, 8150, 8165, 8195, 8215

Offset

1,1

Links

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

Example

6740 is a term because 6740 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4.

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

Crossrefs

Cf. A345556, A345592, A345600, A345602, A345640, A345860.

Sequence in context: A046410 A326811 A098936 * A345860 A206008 A358869

Adjacent sequences: A345598 A345599 A345600 * A345602 A345603 A345604

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved