A345787 Numbers that are the sum of eight cubes in exactly five ways.

471, 497, 504, 597, 623, 630, 635, 642, 649, 654, 661, 667, 680, 686, 691, 693, 712, 717, 723, 728, 736, 738, 741, 743, 752, 754, 755, 762, 774, 780, 781, 783, 784, 785, 788, 791, 793, 797, 800, 802, 804, 810, 813, 814, 815, 817, 819, 820, 821, 830, 834, 837

Offset

1,1

Comments

Differs from A345535 at term 6 because 628 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 + 7^3.

Likely finite.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..180

Example

497 is a term because 497 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 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, 8):
    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. A345535, A345777, A345786, A345788, A345797, A345837.

Sequence in context: A251771 A251152 A345535 * A265137 A265138 A269749

Adjacent sequences: A345784 A345785 A345786 * A345788 A345789 A345790

Keyword

nonn

Author

David Consiglio, Jr., Jun 26 2021

Status

approved