A345816 Numbers that are the sum of six fourth powers in exactly four ways.

6626, 6691, 6866, 9251, 9491, 10115, 10706, 10786, 11555, 12595, 14225, 14691, 14771, 15315, 15330, 15570, 16051, 16595, 16660, 16675, 16850, 17090, 17091, 17236, 17316, 17331, 17346, 17860, 17875, 17940, 17955, 18195, 18786, 18851, 19155, 19170, 19475, 19490

Offset: 1

Author

David Consiglio, Jr., Jun 26 2021

Keywords

• nonn

Comments

Differs from A345561 at term 16 because 15395 = 1^4 + 1^4 + 1^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 5^4 + 8^4 + 8^4 + 9^4 = 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 10^4 = 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 11^4.

Examples

			6691 is a term because 6691 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4.
		

Links

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

Crossrefs

Cf. A344355, A345561, A345766, A345815, A345817, A345826, A346359.

Programs

Mathematica

Select[Range[20000],Count[PowersRepresentations[#,6,4],?(#[[1]]>0&)]==4&] (* _Harvey P. Dale, Mar 11 2023 *)

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 == 4])
    for x in range(len(rets)):
        print(rets[x])