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A345627
Numbers that are the sum of nine fifth powers in ten or more ways.
6
4157156, 4492410, 4510461, 4915538, 4948274, 5005474, 5015506, 5179747, 5219655, 5252477, 5739988, 5756794, 6323426, 6326519, 6382443, 6423394, 6654999, 6705284, 6793170, 6861218, 7101038, 7147645, 7147656, 7148679, 7266240, 7280391, 7283268, 7314187, 7413493
OFFSET
1,1
LINKS
EXAMPLE
4492410 is a term because 4492410 = 1^5 + 1^5 + 2^5 + 3^5 + 5^5 + 7^5 + 7^5 + 13^5 + 21^5 = 1^5 + 2^5 + 6^5 + 10^5 + 11^5 + 11^5 + 14^5 + 16^5 + 19^5 = 1^5 + 6^5 + 7^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 + 18^5 = 2^5 + 5^5 + 6^5 + 6^5 + 7^5 + 15^5 + 15^5 + 16^5 + 18^5 = 2^5 + 5^5 + 6^5 + 10^5 + 10^5 + 11^5 + 11^5 + 15^5 + 20^5 = 3^5 + 3^5 + 7^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 + 18^5 = 3^5 + 3^5 + 8^5 + 8^5 + 8^5 + 12^5 + 12^5 + 17^5 + 19^5 = 3^5 + 4^5 + 6^5 + 7^5 + 8^5 + 13^5 + 14^5 + 16^5 + 19^5 = 4^5 + 4^5 + 4^5 + 7^5 + 11^5 + 11^5 + 13^5 + 18^5 + 18^5 = 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 9^5 + 16^5 + 17^5 + 18^5.
PROG
(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 >= 10])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved