A345566
Numbers that are the sum of six fourth powers in nine or more ways.
Original entry on oeis.org
88595, 122915, 132546, 134931, 144835, 146450, 151556, 161475, 162355, 162755, 170275, 171555, 171795, 172036, 172835, 173075, 177380, 177716, 180770, 183540, 183620, 184835, 185315, 185555, 187700, 187715, 190100, 190211, 193635, 195380, 195780, 196435
Offset: 1
122915 is a term because 122915 = 1^4 + 3^4 + 6^4 + 9^4 + 10^4 + 18^4 = 1^4 + 4^4 + 7^4 + 8^4 + 15^4 + 16^4 = 1^4 + 7^4 + 9^4 + 10^4 + 14^4 + 16^4 = 2^4 + 3^4 + 4^4 + 5^4 + 14^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 11^4 + 18^4 = 2^4 + 9^4 + 9^4 + 12^4 + 14^4 + 15^4 = 3^4 + 5^4 + 6^4 + 6^4 + 11^4 + 18^4 = 3^4 + 8^4 + 10^4 + 11^4 + 13^4 + 16^4 = 5^4 + 6^4 + 7^4 + 11^4 + 14^4 + 16^4 = 8^4 + 8^4 + 9^4 + 10^4 + 11^4 + 17^4.
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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, 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])
A345631
Numbers that are the sum of seven fifth powers in nine or more ways.
Original entry on oeis.org
110276376, 124732805, 127808693, 130298618, 134581976, 188116743, 189642309, 202274051, 202686274, 203343582, 219063107, 230909843, 233137574, 233549568, 234250752, 235438301, 244250335, 251138524, 252277376, 253480833, 254017026, 254380543
Offset: 1
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.
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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])
A345722
Numbers that are the sum of six fifth powers in eight or more ways.
Original entry on oeis.org
2295937600, 4335900525, 6251954544, 8986552608, 9085584992, 13413708308, 14539246326, 15277569450, 15728636000, 16770321920, 16873011232, 16933805856, 17572402769, 17713454592, 17960776999, 18190647200, 19621666592, 20570070125, 20827689300
Offset: 1
4335900525 is a term because 4335900525 = 2^5 + 24^5 + 34^5 + 56^5 + 61^5 + 78^5 = 3^5 + 21^5 + 37^5 + 54^5 + 62^5 + 78^5 = 3^5 + 21^5 + 39^5 + 49^5 + 66^5 + 77^5 = 3^5 + 26^5 + 32^5 + 49^5 + 72^5 + 73^5 = 8^5 + 16^5 + 42^5 + 49^5 + 61^5 + 79^5 = 9^5 + 13^5 + 43^5 + 47^5 + 66^5 + 77^5 = 19^5 + 20^5 + 30^5 + 45^5 + 61^5 + 80^5 = 21^5 + 24^5 + 28^5 + 37^5 + 67^5 + 78^5.
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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 >= 8])
for x in range(len(rets)):
print(rets[x])
A346364
Numbers that are the sum of six fifth powers in exactly nine ways.
Original entry on oeis.org
9085584992, 16933805856, 37377003050, 39254220544, 41066625600, 41485873792, 42149876800, 43828403850, 44180505600, 45902654525, 48588434400, 52005184992, 53536896864, 54156285568, 56229189632, 57088402525, 59954496800, 63432407850, 66188522400, 66507304800
Offset: 1
9085584992 = 24^5 + 38^5 + 42^5 + 48^5 + 54^5 + 96^5
= 21^5 + 34^5 + 38^5 + 43^5 + 74^5 + 92^5
= 8^5 + 34^5 + 38^5 + 62^5 + 68^5 + 92^5
= 18^5 + 18^5 + 44^5 + 64^5 + 66^5 + 92^5
= 13^5 + 18^5 + 51^5 + 64^5 + 64^5 + 92^5
= 8^5 + 38^5 + 41^5 + 47^5 + 79^5 + 89^5
= 5^5 + 23^5 + 29^5 + 45^5 + 85^5 + 85^5
= 8^5 + 23^5 + 41^5 + 64^5 + 82^5 + 84^5
= 12^5 + 18^5 + 38^5 + 72^5 + 78^5 + 84^5,
so 9085584992 is a term.
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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])
A344196
Numbers that are the sum of six fifth powers in ten or more ways.
Original entry on oeis.org
55302546200, 89999127392, 96110537743, 104484239200, 120492759200, 121258798144, 127794946400, 133364991375, 135030535200, 136156575744, 151305014432, 155434423925, 174388570400, 177099008000, 179272687000, 180336745600, 182844944832, 184948721056, 187873845500
Offset: 1
89999127392 = 4^5 + 36^5 + 39^5 + 40^5 + 90^5 + 153^5
= 8^5 + 21^5 + 90^5 + 109^5 + 119^5 + 135^5
= 8^5 + 28^5 + 98^5 + 102^5 + 104^5 + 142^5
= 10^5 + 38^5 + 74^5 + 102^5 + 118^5 + 140^5
= 13^5 + 51^5 + 64^5 + 98^5 + 112^5 + 144^5
= 18^5 + 44^5 + 66^5 + 98^5 + 112^5 + 144^5
= 18^5 + 52^5 + 72^5 + 78^5 + 118^5 + 144^5
= 28^5 + 60^5 + 63^5 + 65^5 + 124^5 + 142^5
= 36^5 + 53^5 + 62^5 + 63^5 + 129^5 + 139^5
= 39^5 + 41^5 + 64^5 + 91^5 + 98^5 + 149^5
so 89999127392 is a term.
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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 >= 10])
for x in range(len(rets)):
print(rets[x])
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