A345818 Numbers that are the sum of six fourth powers in exactly six ways.

37811, 38051, 43251, 43571, 44115, 44531, 45155, 45651, 45891, 47411, 47586, 49971, 52195, 53235, 54131, 56290, 57395, 57460, 57570, 59075, 59330, 59860, 60035, 62180, 62211, 63971, 66340, 67026, 67635, 67715, 67860, 67940, 68115, 68291, 68484, 69395, 69410

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345563 at term 1 because 21251 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4.

Examples

			37811 is a term because 37811 = 1^4 + 2^4 + 2^4 + 7^4 + 11^4 + 12^4 = 2^4 + 2^4 + 4^4 + 7^4 + 9^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 9^4 + 13^4 = 3^4 + 4^4 + 8^4 + 8^4 + 11^4 + 11^4 = 4^4 + 6^4 + 7^4 + 9^4 + 9^4 + 12^4 = 5^4 + 5^4 + 9^4 + 10^4 + 10^4 + 10^4.
		

Links

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

Crossrefs

Cf. A344941, A345563, A345768, A345817, A345819, A345828, A346361.

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, 6):
    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])