A345583 Numbers that are the sum of eight fourth powers in eight or more ways.

13268, 14212, 14788, 15427, 15667, 16612, 16627, 16692, 16707, 16772, 16822, 16852, 16882, 16947, 17348, 17363, 17428, 17493, 17877, 17972, 17987, 18052, 18117, 18227, 18948, 19157, 19237, 19252, 19267, 19412, 19492, 19507, 19572, 19682, 19747, 19748, 19828

Offset

1,1

Links

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

Example

14212 is a term because 14212 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 8^4 + 10^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 10^4 = 1^4 + 1^4 + 1^4 + 5^4 + 6^4 + 8^4 + 8^4 + 8^4 = 1^4 + 2^4 + 4^4 + 4^4 + 5^4 + 7^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 8^4 + 9^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 10^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 10^4 = 3^4 + 4^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 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, 8):
    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. A345538, A345574, A345582, A345584, A345592, A345616, A345840.

Sequence in context: A346904 A235967 A184001 * A345840 A031620 A251339

Adjacent sequences: A345580 A345581 A345582 * A345584 A345585 A345586

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved