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A345794
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Numbers that are the sum of nine cubes in exactly two ways.
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7
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72, 133, 140, 147, 159, 161, 166, 168, 175, 182, 185, 187, 189, 194, 196, 198, 201, 203, 205, 208, 213, 217, 220, 222, 227, 239, 243, 246, 252, 261, 265, 266, 273, 289, 296, 304, 306, 308, 323, 325, 328, 329, 330, 336, 342, 344, 349, 351, 352, 354, 356, 358
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OFFSET
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1,1
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COMMENTS
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Differs from A345541 at term 25 because 224 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.
Likely finite.
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LINKS
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EXAMPLE
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133 is a term because 133 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3.
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PROG
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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, 9):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 2])
for x in range(len(rets)):
print(rets[x])
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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