A345555 Numbers that are the sum of ten cubes in seven or more ways.

440, 473, 499, 506, 525, 532, 534, 567, 571, 584, 588, 597, 599, 604, 606, 623, 625, 627, 630, 632, 637, 639, 640, 644, 651, 656, 658, 660, 662, 663, 665, 669, 670, 673, 677, 680, 682, 684, 688, 689, 691, 693, 695, 696, 697, 699, 701, 702, 704, 707, 708, 714

Offset

1,1

Links

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

Example

473 is a term because 473 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

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 >= 7])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345546, A345554, A345556, A345600, A345809, A346806.

Sequence in context: A210205 A092048 A296905 * A345809 A279812 A279950

Adjacent sequences: A345552 A345553 A345554 * A345556 A345557 A345558

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved