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A345624
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Numbers that are the sum of nine fifth powers in seven or more ways.
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7
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1431398, 1431429, 1431640, 1439173, 1447570, 1504636, 1531397, 1597929, 1671167, 1696159, 1697686, 1697928, 1778835, 1936454, 1952415, 1969221, 1975049, 2017344, 2092122, 2182161, 2198967, 2208680, 2247917, 2280818, 2283911, 2289343, 2314335, 2329845, 2340319
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OFFSET
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1,1
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LINKS
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EXAMPLE
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1431429 is a term because 1431429 = 1^5 + 2^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 14^5 + 15^5 = 2^5 + 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 2^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 2^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5.
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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, 9):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 7])
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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