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A346329
Numbers that are the sum of eight fifth powers in exactly four ways.
7
391250, 392031, 455750, 519236, 604822, 622281, 672023, 672054, 672265, 673554, 697492, 703978, 707368, 730259, 763292, 857761, 893605, 893636, 893816, 893847, 894027, 894058, 894452, 894628, 896729, 897151, 901380, 903839, 909124, 909597, 910411, 911403
OFFSET
1,1
COMMENTS
Differs from A345612 at term 33 because 926372 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 10^5 + 15^5 = 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 10^5 + 15^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 6^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5.
LINKS
EXAMPLE
391250 is a term because 391250 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5.
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, 8):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 4])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved