| Displaying 1-6 of 6 results found.
page
1
Numbers that are the sum of four third powers in eight or more ways.
+10
9
21896, 27720, 30429, 31339, 31402, 33579, 34624, 34776, 36162, 36225, 40105, 42120, 42695, 44037, 44163, 44226, 44947, 45162, 45675, 46277, 46683, 46872, 46900, 47600, 48321, 48825, 49042, 50112, 50689, 50806, 50904, 51058, 51408, 51480, 51506, 51597, 51688
EXAMPLE
30429 is a term because 30429 = 1^3 + 4^3 + 7^3 + 30^3 = 1^3 + 16^3 + 17^3 + 26^3 = 2^3 + 12^3 + 21^3 + 25^3 = 3^3 + 3^3 + 14^3 + 29^3 = 4^3 + 17^3 + 21^3 + 23^3 = 5^3 + 11^3 + 15^3 + 28^3 = 6^3 + 6^3 + 22^3 + 25^3 = 7^3 + 14^3 + 18^3 + 26^3.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 4):
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])
Numbers that are the sum of four fourth powers in exactly eight ways.
+10
7
13155858, 26421474, 35965458, 39803778, 98926434, 128198994, 143776179, 156279618, 210493728, 237073554, 248075538, 255831858, 257931378, 269965938, 270289698, 292967619, 293579874, 295880274, 300120003, 301080243, 302115843, 305670834, 309742434, 331957458
COMMENTS
Differs from A344924 at term 24 because 328118259 = 2^4 + 77^4 + 109^4 + 111^4 = 8^4 + 79^4 + 93^4 + 121^4 = 18^4 + 79^4 + 97^4 + 119^4 = 21^4 + 77^4 + 98^4 + 119^4 = 27^4 + 77^4 + 94^4 + 121^4 = 34^4 + 77^4 + 89^4 + 123^4 = 46^4 + 57^4 + 103^4 + 119^4 = 49^4 + 77^4 + 77^4 + 126^4 = 61^4 + 66^4 + 77^4 + 127^4.
EXAMPLE
13155858 is a term because 13155858 = 1^4 + 16^4 + 19^4 + 60^4 = 3^4 + 6^4 + 21^4 + 60^4 = 10^4 + 18^4 + 31^4 + 59^4 = 12^4 + 27^4 + 45^4 + 54^4 = 15^4 + 44^4 + 46^4 + 47^4 = 18^4 + 25^4 + 41^4 + 56^4 = 29^4 + 30^4 + 44^4 + 53^4 = 35^4 + 36^4 + 38^4 + 53^4.
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, 4):
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])
Numbers that are the sum of three third powers in exactly eight ways.
+10
7
2562624, 5618250, 6525000, 6755328, 7374375, 12742920, 13581352, 14027112, 14288373, 18443160, 20500992, 22783032, 23113728, 25305048, 26936064, 27131840, 29515968, 30205440, 32835375, 38269440, 39317832, 39339000, 40189248, 42144192, 42183504, 43077952
COMMENTS
Differs from A345087 at term 10 because 14926248 = 2^3 + 33^3 + 245^3 = 11^3 + 185^3 + 203^3 = 14^3 + 32^3 + 245^3 = 50^3 + 113^3 + 236^3 = 71^3 + 89^3 + 239^3 = 74^3 + 189^3 + 196^3 = 89^3 + 185^3 + 197^3 = 98^3 + 148^3 + 219^3 = 105^3 + 149^3 + 217^3.
EXAMPLE
2562624 is a term because 2562624 = 7^3 + 35^3 + 135^3 = 7^3 + 63^3 + 131^3 = 11^3 + 99^3 + 115^3 = 16^3 + 45^3 + 134^3 = 29^3 + 102^3 + 112^3 = 35^3 + 59^3 + 131^3 = 50^3 + 84^3 + 121^3 = 68^3 + 71^3 + 122^3.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 3):
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])
Numbers that are the sum of four third powers in exactly seven ways.
+10
7
13104, 18928, 19376, 20755, 21203, 22743, 24544, 24570, 24787, 25172, 25928, 27755, 27846, 28917, 29582, 31031, 31248, 31528, 32858, 34056, 34713, 35289, 35317, 35441, 35497, 35712, 36190, 36288, 36610, 36890, 36946, 38080, 39221, 39440, 39464, 39851, 39942
COMMENTS
Differs from A345150 at term 6 because 21896 = 1^3 + 11^3 + 19^3 + 22^3 = 2^3 + 2^3 + 12^3 + 26^3 = 2^3 + 3^3 + 19^3 + 23^3 = 2^3 + 5^3 + 15^3 + 25^3 = 3^3 + 10^3 + 16^3 + 24^3 = 3^3 + 17^3 + 19^3 + 19^3 = 4^3 + 6^3 + 20^3 + 22^3 = 5^3 + 8^3 + 14^3 + 25^3 = 7^3 + 11^3 + 17^3 + 23^3 = 8^3 + 9^3 + 19^3 + 22^3.
EXAMPLE
13104 is a term because 13104 = 1^3 + 10^3 + 16^3 + 18^3 = 1^3 + 11^3 + 14^3 + 19^3 = 2^3 + 9^3 + 15^3 + 19^3 = 4^3 + 6^3 + 14^3 + 20^3 = 4^3 + 9^3 + 10^3 + 21^3 = 5^3 + 7^3 + 11^3 + 21^3 = 8^3 + 9^3 + 14^3 + 19^3.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 4):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 7])
for x in range(len(rets)):
print(rets[x])
Numbers that are the sum of four third powers in exactly nine ways.
+10
7
42120, 46683, 50806, 50904, 51408, 51480, 51688, 52208, 53865, 54971, 56385, 57113, 60515, 60984, 62433, 65303, 66276, 66339, 66430, 67158, 69048, 69832, 69930, 71162, 72072, 72520, 72576, 72800, 73017, 77714, 77903, 79345, 79667, 79849, 80066, 80073, 81207
COMMENTS
Differs from A345146 at term 1 because 21896 = 1^3 + 11^3 + 19^3 + 22^3 = 2^3 + 2^3 + 12^3 + 26^3 = 2^3 + 3^3 + 19^3 + 23^3 = 2^3 + 5^3 + 15^3 + 25^3 = 3^3 + 10^3 + 16^3 + 24^3 = 3^3 + 17^3 + 19^3 + 19^3 = 4^3 + 6^3 + 20^3 + 22^3 = 5^3 + 8^3 + 14^3 + 25^3 = 7^3 + 11^3 + 17^3 + 23^3 = 8^3 + 9^3 + 19^3 + 22^3.
EXAMPLE
42120 is a term because 42120 = 1^3 + 19^3 + 22^3 + 27^3 = 2^3 + 3^3 + 13^3 + 33^3 = 2^3 + 6^3 + 17^3 + 32^3 = 3^3 + 3^3 + 20^3 + 31^3 = 3^3 + 17^3 + 20^3 + 29^3 = 3^3 + 13^3 + 14^3 + 32^3 = 6^3 + 15^3 + 16^3 + 31^3 = 7^3 + 17^3 + 23^3 + 27^3 = 11^3 + 13^3 + 21^3 + 29^3.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 4):
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])
Numbers that are the sum of five third powers in exactly eight ways.
+10
7
4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5914, 6075, 6202, 6499, 6560, 6616, 6642, 6677, 6833, 6884, 7008, 7111, 7128, 7155, 7218, 7344, 7395, 7641, 7696, 7729, 7785, 7813, 7820, 7849, 7883, 8037, 8100, 8243, 8282, 8308, 8315
COMMENTS
Differs from A345183 at term 13 because 5860 = 1^3 + 1^3 + 5^3 + 8^3 + 16^3 = 1^3 + 2^3 + 3^3 + 11^3 + 15^3 = 1^3 + 3^3 + 8^3 + 11^3 + 14^3 = 1^3 + 5^3 + 5^3 + 10^3 + 15^3 = 1^3 + 9^3 + 10^3 + 10^3 + 12^3 = 2^3 + 3^3 + 8^3 + 9^3 + 15^3 = 2^3 + 3^3 + 5^3 + 12^3 + 14^3 = 2^3 + 8^3 + 8^3 + 12^3 + 12^3 = 3^3 + 8^3 + 8^3 + 9^3 + 14^3 = 3^3 + 6^3 + 7^3 + 12^3 + 13^3.
EXAMPLE
4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3 = 1^3 + 3^3 + 7^3 + 9^3 + 14^3 = 1^3 + 8^3 + 8^3 + 11^3 + 11^3 = 2^3 + 4^3 + 6^3 + 6^3 + 15^3 = 3^3 + 3^3 + 5^3 + 7^3 + 15^3 = 3^3 + 3^3 + 10^3 + 11^3 + 11^3 = 4^3 + 6^3 + 6^3 + 8^3 + 14^3 = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 5):
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])
Search completed in 0.007 seconds
|