|
|
A345723
|
|
Numbers that are the sum of six fifth powers in nine or more ways.
|
|
6
|
|
|
9085584992, 16933805856, 37377003050, 39254220544, 41066625600, 41485873792, 42149876800, 43828403850, 44180505600, 45902654525, 48588434400, 52005184992, 53536896864, 54156285568, 55302546200, 56229189632, 57088402525, 59954496800, 63432407850
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
LINKS
|
|
|
EXAMPLE
|
16933805856 is a term because 16933805856 = 2^5 + 38^5 + 68^5 + 74^5 + 92^5 + 92^5 = 2^5 + 54^5 + 58^5 + 64^5 + 92^5 + 96^5 = 14^5 + 36^5 + 61^5 + 67^5 + 94^5 + 94^5 = 15^5 + 49^5 + 52^5 + 60^5 + 94^5 + 96^5 = 17^5 + 49^5 + 53^5 + 57^5 + 92^5 + 98^5 = 29^5 + 36^5 + 42^5 + 72^5 + 88^5 + 99^5 = 31^5 + 36^5 + 54^5 + 54^5 + 94^5 + 97^5 = 34^5 + 34^5 + 46^5 + 72^5 + 76^5 + 104^5 = 35^5 + 36^5 + 69^5 + 72^5 + 89^5 + 95^5.
|
|
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, 6):
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
|
|
|
|