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A345599 Numbers that are the sum of ten fourth powers in six or more ways. 7
3175, 4150, 4230, 4390, 4405, 4455, 4470, 4485, 4500, 4550, 4565, 4630, 4725, 4740, 4915, 4980, 5094, 5109, 5155, 5190, 5205, 5220, 5270, 5285, 5350, 5365, 5395, 5430, 5445, 5460, 5475, 5525, 5540, 5590, 5605, 5635, 5655, 5670, 5700, 5715, 5735, 5765, 5780 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
LINKS
EXAMPLE
4150 is a term because 4150 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 = 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 7^4.
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, 10):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 6])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
Sequence in context: A217247 A020420 A155484 * A345858 A129472 A023309
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified February 22 22:43 EST 2024. Contains 370265 sequences. (Running on oeis4.)