A345808 Numbers that are the sum of ten cubes in exactly six ways.

436, 447, 466, 477, 480, 492, 503, 508, 510, 513, 515, 518, 527, 529, 536, 538, 539, 541, 551, 553, 560, 562, 564, 569, 577, 581, 590, 595, 601, 602, 603, 607, 608, 613, 614, 616, 618, 621, 628, 634, 636, 642, 643, 645, 647, 649, 654, 655, 659, 666, 678, 679

Offset

1,1

Comments

Differs from A345554 at term 2 because 440 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..73

Example

440 is a term because 440 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.

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, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 6])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345554, A345798, A345807, A345809, A345858.

Sequence in context: A237302 A124043 A345554 * A260366 A250781 A260012

Adjacent sequences: A345805 A345806 A345807 * A345809 A345810 A345811

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved