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A345575
Numbers that are the sum of seven fourth powers in nine or more ways.
8
19491, 21267, 21332, 23652, 31251, 35427, 36052, 37812, 38067, 39891, 40356, 41732, 41747, 43267, 43876, 43891, 43956, 44131, 44196, 44532, 44547, 44612, 45156, 45171, 45411, 45651, 45652, 45827, 45891, 45892, 45907, 46276, 46451, 46516, 47427, 47667, 47971
OFFSET
1,1
LINKS
EXAMPLE
21267 is a term because 21267 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 2^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1, 1000)]
for pos in cwr(power_terms, 7):
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])
KEYWORD
nonn
AUTHOR
STATUS
approved