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A346348
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Numbers that are the sum of ten fifth powers in exactly three ways.
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6
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8194, 21940, 52419, 52450, 52481, 52661, 52692, 52903, 53442, 53473, 53684, 54465, 54520, 54551, 54582, 54762, 54793, 55004, 55691, 55722, 55933, 56714, 57644, 57675, 57886, 58815, 60194, 60225, 60436, 60768, 61217, 62295, 62326, 62537, 63466, 65419, 67969
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OFFSET
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1,1
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COMMENTS
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Differs from A345635 at term 19 because 55543 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5.
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LINKS
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EXAMPLE
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8194 is a term because 8194 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5.
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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**5 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 == 3])
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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