A346281 Numbers that are the sum of seven fifth powers in exactly four ways.

893604, 1117071, 1182534, 1414559, 1629244, 1933328, 2280543, 2586035, 2867074, 3050644, 3055295, 3055977, 3256432, 3329360, 3369543, 3436058, 3551890, 3576363, 3896969, 3914877, 3930849, 4055954, 4087746, 4088690, 4093572, 4096665, 4098161, 4104068, 4104310

Offset

1,1

Comments

Differs from A345607 at term 92 because 6768576 = 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 12^5 + 23^5 = 1^5 + 3^5 + 4^5 + 8^5 + 11^5 + 17^5 + 22^5 = 6^5 + 12^5 + 13^5 + 14^5 + 15^5 + 15^5 + 21^5 = 8^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 20^5 = 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 20^5.

Links

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

Example

893604 is a term because 893604 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 15^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5.

Prog

(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**5 for x in range(1, 1000)]
for pos in cwr(power_terms, 7):
    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. A345607, A345826, A346280, A346282, A346329, A346359.

Sequence in context: A105379 A186792 A345607 * A032429 A184456 A190391

Adjacent sequences: A346278 A346279 A346280 * A346282 A346283 A346284

Keyword

nonn

Author

David Consiglio, Jr., Jul 13 2021

Status

approved