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A345631
Numbers that are the sum of seven fifth powers in nine or more ways.
7
110276376, 124732805, 127808693, 130298618, 134581976, 188116743, 189642309, 202274051, 202686274, 203343582, 219063107, 230909843, 233137574, 233549568, 234250752, 235438301, 244250335, 251138524, 252277376, 253480833, 254017026, 254380543
OFFSET
1,1
LINKS
EXAMPLE
124732805 is a term because 124732805 = 3^5 + 18^5 + 22^5 + 22^5 + 24^5 + 27^5 + 39^5 = 4^5 + 15^5 + 17^5 + 21^5 + 29^5 + 34^5 + 35^5 = 5^5 + 14^5 + 17^5 + 24^5 + 25^5 + 35^5 + 35^5 = 7^5 + 8^5 + 17^5 + 26^5 + 29^5 + 34^5 + 34^5 = 7^5 + 10^5 + 12^5 + 31^5 + 31^5 + 32^5 + 32^5 = 7^5 + 12^5 + 23^5 + 24^5 + 24^5 + 26^5 + 39^5 = 7^5 + 14^5 + 22^5 + 22^5 + 23^5 + 28^5 + 39^5 = 16^5 + 25^5 + 25^5 + 28^5 + 28^5 + 28^5 + 35^5 = 20^5 + 24^5 + 24^5 + 25^5 + 25^5 + 32^5 + 35^5.
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, 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])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved