A345553 Numbers that are the sum of ten cubes in five or more ways.
288, 349, 382, 384, 401, 403, 408, 410, 414, 415, 417, 421, 429, 436, 440, 443, 447, 454, 455, 462, 466, 473, 475, 477, 480, 482, 487, 492, 496, 499, 501, 503, 506, 508, 510, 513, 515, 518, 520, 525, 527, 529, 532, 534, 536, 538, 539, 541, 543, 544, 545, 546
Offset: 1
Keywords
Examples
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.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..10000
Programs
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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])