A345817 Numbers that are the sum of six fourth powers in exactly five ways.

15395, 16610, 18866, 19235, 19410, 20996, 21011, 21316, 21331, 21491, 21620, 23811, 25091, 29700, 29715, 29906, 29955, 30356, 30995, 31235, 31266, 31331, 31506, 32035, 33651, 33795, 33891, 35171, 35411, 35636, 35796, 35971, 37971, 38595, 38675, 39266, 39890

Offset

1,1

Comments

Differs from A345562 at term 8 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.

Links

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

Example

16610 is a term because 16610 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 5^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 10^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4.

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

Crossrefs

Cf. A344359, A345562, A345767, A345816, A345818, A345827, A346360.

Sequence in context: A362880 A054837 A345562 * A252990 A252842 A261724

Adjacent sequences: A345814 A345815 A345816 * A345818 A345819 A345820

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved