A345554 Numbers that are the sum of ten cubes in six or more ways.
436, 440, 447, 466, 473, 477, 480, 492, 499, 503, 506, 508, 510, 513, 515, 518, 525, 527, 529, 532, 534, 536, 538, 539, 541, 551, 553, 560, 562, 564, 567, 569, 571, 577, 581, 584, 588, 590, 595, 597, 599, 601, 602, 603, 604, 606, 607, 608, 613, 614, 616, 618
Offset: 1
Keywords
Examples
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.
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 >= 6]) for x in range(len(rets)): print(rets[x])