A345186 Numbers that are the sum of five third powers in exactly nine ways.

6112, 6138, 6462, 6497, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8163, 8352, 8443, 8560, 8630, 8632, 8928, 8991, 9017, 9136, 9143, 9171, 9288, 9316, 9379, 9505, 9566, 9647, 9658, 9675, 9684, 9745, 9773

Offset

1,1

Comments

Differs from A345185 at term 1 because 5860 = 1^3 + 1^3 + 5^3 + 8^3 + 16^3 = 1^3 + 2^3 + 3^3 + 11^3 + 15^3 = 1^3 + 3^3 + 8^3 + 11^3 + 14^3 = 1^3 + 5^3 + 5^3 + 10^3 + 15^3 = 1^3 + 9^3 + 10^3 + 10^3 + 12^3 = 2^3 + 3^3 + 8^3 + 9^3 + 15^3 = 2^3 + 3^3 + 5^3 + 12^3 + 14^3 = 2^3 + 8^3 + 8^3 + 12^3 + 12^3 = 3^3 + 8^3 + 8^3 + 9^3 + 14^3 = 3^3 + 6^3 + 7^3 + 12^3 + 13^3.

Links

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Example

6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3  = 1^3 + 3^3 + 7^3 + 12^3 + 14^3  = 1^3 + 6^3 + 6^3 + 7^3 + 16^3  = 2^3 + 2^3 + 9^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 11^3 + 15^3  = 2^3 + 8^3 + 9^3 + 9^3 + 14^3  = 3^3 + 3^3 + 3^3 + 4^3 + 17^3  = 3^3 + 5^3 + 8^3 + 11^3 + 14^3  = 8^3 + 8^3 + 8^3 + 11^3 + 12^3.

Prog

(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 9])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A294743, A341892, A345154, A345184, A345185, A345188, A345771.

Sequence in context: A031619 A235764 A235547 * A251368 A366516 A224483

Adjacent sequences: A345183 A345184 A345185 * A345187 A345188 A345189

Keyword

nonn

Author

David Consiglio, Jr., Jun 10 2021

Status

approved