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

1,1

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.

Links

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

Example

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.

Mathematica

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

Prog

(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])

Crossrefs

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

Sequence in context: A237320 A237726 A345561 * A250539 A251796 A237794

Adjacent sequences: A345813 A345814 A345815 * A345817 A345818 A345819

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved