A345826 Numbers that are the sum of seven fourth powers in exactly four ways.

2932, 4147, 4212, 4387, 5427, 5602, 5667, 6627, 6692, 6817, 6822, 6837, 6852, 6867, 7012, 7122, 7251, 7316, 7491, 7747, 7857, 8052, 8097, 8162, 8402, 8467, 8532, 8707, 8787, 9027, 9092, 9157, 9172, 9202, 9237, 9252, 9332, 9412, 9442, 9492, 9572, 9652, 9682

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345570 at term 9 because 6642 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Examples

			4147 is a term because 4147 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4.
		

Links

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

Crossrefs

Cf. A345570, A345776, A345816, A345825, A345827, A345836, A346281.

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, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 4])
    for x in range(len(rets)):
        print(rets[x])