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A345787
Numbers that are the sum of eight cubes in exactly five ways.
7
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
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
KEYWORD
nonn
AUTHOR
STATUS
approved