A345807 Numbers that are the sum of ten cubes in exactly five ways.

288, 349, 382, 384, 401, 403, 408, 410, 414, 415, 417, 421, 429, 443, 454, 455, 462, 475, 482, 487, 496, 501, 520, 543, 544, 545, 546, 548, 555, 556, 558, 565, 566, 570, 573, 574, 575, 576, 578, 580, 582, 586, 587, 591, 592, 593, 594, 600, 609, 610, 611, 617

Offset

1,1

Comments

Differs from A345553 at term 14 because 436 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..86

Example

349 is a term because 349 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.

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

Crossrefs

Cf. A345553, A345797, A345806, A345808, A345857.

Sequence in context: A371948 A287116 A345553 * A380937 A244196 A392134

Adjacent sequences: A345804 A345805 A345806 * A345808 A345809 A345810

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved