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A345626
Numbers that are the sum of nine fifth powers in nine or more ways.
7
1969221, 2596936, 3353186, 3378178, 3923426, 3981447, 4094027, 4096729, 4112329, 4114188, 4129465, 4137209, 4147736, 4157156, 4170112, 4172994, 4254304, 4303773, 4410482, 4475846, 4477936, 4483379, 4485480, 4492410, 4501441, 4510461, 4543232, 4652011, 4691855
OFFSET
1,1
LINKS
EXAMPLE
2596936 is a term because 2596936 = 1^5 + 1^5 + 4^5 + 5^5 + 9^5 + 13^5 + 13^5 + 13^5 + 17^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 17^5 = 1^5 + 4^5 + 7^5 + 7^5 + 7^5 + 9^5 + 9^5 + 14^5 + 18^5 = 1^5 + 5^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 12^5 + 13^5 + 14^5 + 17^5 = 2^5 + 5^5 + 5^5 + 6^5 + 6^5 + 6^5 + 15^5 + 15^5 + 16^5 = 3^5 + 3^5 + 5^5 + 6^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 11^5 + 11^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 8^5 + 9^5 + 16^5 + 17^5.
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 >= 9])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved