A345821
Numbers that are the sum of six fourth powers in exactly nine ways.
Original entry on oeis.org
88595, 132546, 134931, 144835, 146450, 162355, 170275, 171555, 171795, 172036, 172835, 177380, 177716, 180770, 183540, 184835, 185555, 187700, 187715, 190100, 190211, 193635, 195380, 195780, 196435, 197780, 199075, 199475, 199730, 199955, 202196, 202980
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.
-
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])
A346286
Numbers that are the sum of seven fifth powers in exactly nine ways.
Original entry on oeis.org
110276376, 124732805, 127808693, 130298618, 188116743, 202274051, 202686274, 203343582, 230909843, 233137574, 233549568, 234250752, 244250335, 251138524, 253480833, 254017026, 254380543, 265006057, 265072501, 273628068, 279536432, 279770326, 280361082
Offset: 1
110276376 is a term because 110276376 = 1^5 + 3^5 + 5^5 + 7^5 + 17^5 + 23^5 + 40^5 = 5^5 + 10^5 + 16^5 + 16^5 + 19^5 + 20^5 + 40^5 = 1^5 + 8^5 + 14^5 + 16^5 + 21^5 + 27^5 + 39^5 = 7^5 + 8^5 + 11^5 + 14^5 + 16^5 + 33^5 + 37^5 = 4^5 + 7^5 + 8^5 + 13^5 + 26^5 + 31^5 + 37^5 = 1^5 + 5^5 + 6^5 + 20^5 + 28^5 + 29^5 + 37^5 = 3^5 + 3^5 + 7^5 + 18^5 + 27^5 + 32^5 + 36^5 = 6^5 + 12^5 + 18^5 + 25^5 + 30^5 + 31^5 + 34^5 = 6^5 + 10^5 + 20^5 + 27^5 + 27^5 + 33^5 + 33^5.
-
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])
A345723
Numbers that are the sum of six fifth powers in nine or more ways.
Original entry on oeis.org
9085584992, 16933805856, 37377003050, 39254220544, 41066625600, 41485873792, 42149876800, 43828403850, 44180505600, 45902654525, 48588434400, 52005184992, 53536896864, 54156285568, 55302546200, 56229189632, 57088402525, 59954496800, 63432407850
Offset: 1
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
so 16933805856 is a term.
-
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])
A346363
Numbers that are the sum of six fifth powers in exactly eight ways.
Original entry on oeis.org
2295937600, 4335900525, 6251954544, 8986552608, 13413708308, 14539246326, 15277569450, 15728636000, 16770321920, 16873011232, 17572402769, 17713454592, 17960776999, 18190647200, 19621666592, 20570070125, 20827689300, 22322555200, 23461554774, 23613244800
Offset: 1
2295937600 = 4^5 + 21^5 + 38^5 + 42^5 + 43^5 + 72^5
= 8^5 + 16^5 + 30^5 + 42^5 + 54^5 + 70^5
= 8^5 + 13^5 + 36^5 + 37^5 + 57^5 + 69^5
= 14^5 + 16^5 + 16^5 + 52^5 + 54^5 + 68^5
= 3^5 + 14^5 + 32^5 + 44^5 + 61^5 + 66^5
= 4^5 + 18^5 + 22^5 + 52^5 + 58^5 + 66^5
= 10^5 + 14^5 + 26^5 + 42^5 + 63^5 + 65^5
= 1^5 + 7^5 + 34^5 + 57^5 + 58^5 + 63^5,
so 2295937600 is a term.
-
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])
A346365
Numbers that are the sum of six fifth powers in exactly ten ways.
Original entry on oeis.org
55302546200, 89999127392, 96110537743, 104484239200, 120492759200, 121258798144, 127794946400, 133364991375, 135030535200, 136156575744, 151305014432, 155434423925, 174388570400, 177099008000, 179272687000, 182844944832, 184948721056, 187873845500
Offset: 1
55302546200 = 34^5 + 38^5 + 50^5 + 57^5 + 95^5 + 136^5
= 23^5 + 49^5 + 61^5 + 69^5 + 107^5 + 131^5
= 24^5 + 37^5 + 63^5 + 81^5 + 104^5 + 131^5
= 21^5 + 35^5 + 60^5 + 94^5 + 100^5 + 130^5
= 57^5 + 60^5 + 71^5 + 75^5 + 109^5 + 128^5
= 19^5 + 37^5 + 56^5 + 96^5 + 104^5 + 128^5
= 35^5 + 41^5 + 53^5 + 69^5 + 115^5 + 127^5
= 16^5 + 49^5 + 53^5 + 83^5 + 112^5 + 127^5
= 35^5 + 37^5 + 40^5 + 88^5 + 119^5 + 121^5
= 11^5 + 24^5 + 71^5 + 104^5 + 109^5 + 121^5
so 55302546200 is a term.
-
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])
Showing 1-5 of 5 results.
Comments