A345837 Numbers that are the sum of eight fourth powers in exactly five ways.

4228, 4403, 4468, 5443, 5508, 5683, 6613, 6643, 6658, 6708, 6773, 6838, 6868, 6883, 6948, 7013, 7093, 7138, 7203, 7267, 7268, 7332, 7397, 7478, 7507, 7572, 7588, 7828, 7858, 7923, 7988, 8113, 8133, 8228, 8353, 8418, 8533, 8547, 8548, 8612, 8723, 8788, 8852

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345580 at term 11 because 6723 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Examples

			4403 is a term because 4403 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.
		

Links

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

Crossrefs

Cf. A345580, A345787, A345827, A345836, A345838, A345847, A346330.

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 == 5])
    for x in range(len(rets)):
        print(rets[x])