A345581 Numbers that are the sum of eight fourth powers in six or more ways.

6723, 6788, 6853, 6898, 6963, 7028, 7938, 8003, 8068, 8178, 8243, 8308, 8483, 8963, 9043, 9173, 9218, 9283, 9348, 9413, 9493, 9523, 9668, 9763, 9828, 10003, 10132, 10258, 10277, 10307, 10372, 10628, 10708, 10738, 10788, 10803, 10868, 10933, 10948, 10978

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			6788 is a term because 6788 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.
		

Links

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

Crossrefs

Cf. A345536, A345572, A345580, A345582, A345590, A345614, A345838.

Programs

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])