A345567 Numbers that are the sum of six fourth powers in ten or more ways.

122915, 151556, 161475, 162755, 173075, 183620, 185315, 197795, 199106, 199940, 201875, 201955, 202275, 204275, 204340, 204595, 206115, 207395, 209795, 211075, 212420, 213731, 217620, 217826, 217891, 218515, 221250, 223715, 223955, 224180, 224451, 225875

Offset

1,1

Links

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

Example

151556 is a term because 151556 = 1^4 + 2^4 + 2^4 + 9^4 + 11^4 + 19^4 = 1^4 + 2^4 + 3^4 + 7^4 + 16^4 + 17^4 = 1^4 + 8^4 + 11^4 + 12^4 + 13^4 + 17^4 = 2^4 + 3^4 + 7^4 + 8^4 + 11^4 + 19^4 = 3^4 + 3^4 + 3^4 + 4^4 + 12^4 + 19^4 = 3^4 + 4^4 + 11^4 + 11^4 + 14^4 + 17^4 = 3^4 + 4^4 + 13^4 + 13^4 + 13^4 + 16^4 = 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 19^4 = 4^4 + 7^4 + 11^4 + 11^4 + 11^4 + 18^4 = 4^4 + 8^4 + 9^4 + 13^4 + 13^4 + 17^4.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A341897, A344196, A345519, A345566, A345576, A345822.

Sequence in context: A236797 A115545 A224584 * A345822 A254017 A254024

Adjacent sequences: A345564 A345565 A345566 * A345568 A345569 A345570

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved