A341892
Numbers that are the sum of five fourth powers in exactly nine ways.
Original entry on oeis.org
619090, 775714, 1100579, 1179379, 1186834, 1243699, 1357315, 1367539, 1373859, 1422595, 1431234, 1436419, 1511299, 1536019, 1699234, 1734899, 1839874, 1858594, 1880850, 1950355, 1951650, 1978915, 2044819, 2052899, 2069955, 2085139, 2101779, 2119459, 2133234
Offset: 1
619090 = 1^4 + 2^4 + 18^4 + 22^4 + 23^4
= 1^4 + 3^4 + 4^4 + 8^4 + 28^4
= 1^4 + 11^4 + 14^4 + 22^4 + 24^4
= 2^4 + 2^4 + 8^4 + 17^4 + 27^4
= 2^4 + 13^4 + 13^4 + 18^4 + 26^4
= 3^4 + 6^4 + 12^4 + 16^4 + 27^4
= 4^4 + 12^4 + 14^4 + 23^4 + 23^4
= 9^4 + 12^4 + 16^4 + 21^4 + 24^4
= 14^4 + 16^4 + 18^4 + 19^4 + 23^4
so 619090 is a term.
-
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, 5):
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])
A345822
Numbers that are the sum of six fourth powers in exactly ten ways.
Original entry on oeis.org
122915, 151556, 161475, 162755, 173075, 183620, 185315, 199106, 199940, 201875, 202275, 204275, 204340, 204595, 206115, 207395, 209795, 211075, 213731, 217826, 217891, 218515, 221250, 223955, 224180, 225875, 226595, 227186, 228035, 236195, 237796, 237890
Offset: 1
151556 is a term because 151556 = 1^4 + 2^4 + 2^4 + 9^4 + 11^4 + 19^4 = 1^4 + 2^4 + 3^4 + 7^4 + 16^4 + 17^4 = 1^4 + 8^4 + 11^4 + 12^4 + 13^4 + 17^4 = 2^4 + 3^4 + 7^4 + 8^4 + 11^4 + 19^4 = 3^4 + 3^4 + 3^4 + 4^4 + 12^4 + 19^4 = 3^4 + 4^4 + 11^4 + 11^4 + 14^4 + 17^4 = 3^4 + 4^4 + 13^4 + 13^4 + 13^4 + 16^4 = 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 19^4 = 4^4 + 7^4 + 11^4 + 11^4 + 11^4 + 18^4 = 4^4 + 8^4 + 9^4 + 13^4 + 13^4 + 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 == 10])
for x in range(len(rets)):
print(rets[x])
A341897
Numbers that are the sum of five fourth powers in ten or more ways.
Original entry on oeis.org
954979, 1205539, 1574850, 1713859, 1801459, 1863859, 1877394, 1882579, 2071939, 2109730, 2138419, 2142594, 2157874, 2225859, 2288179, 2419954, 2492434, 2495939, 2605314, 2663539, 2711394, 2784499, 2835939, 2847394, 2849859, 2880994, 2919154, 2924674, 3007474
Offset: 1
954979 = 1^4 + 2^4 + 11^4 + 19^4 + 30^4
= 1^4 + 7^4 + 18^4 + 25^4 + 26^4
= 3^4 + 8^4 + 17^4 + 20^4 + 29^4
= 4^4 + 8^4 + 13^4 + 25^4 + 27^4
= 4^4 + 9^4 + 10^4 + 11^4 + 31^4
= 6^4 + 6^4 + 15^4 + 21^4 + 29^4
= 7^4 + 10^4 + 18^4 + 19^4 + 29^4
= 11^4 + 11^4 + 20^4 + 22^4 + 27^4
= 16^4 + 17^4 + 17^4 + 24^4 + 25^4
= 18^4 + 19^4 + 20^4 + 23^4 + 23^4
so 954979 is a term.
-
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, 5):
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])
A344929
Numbers that are the sum of four fourth powers in exactly ten ways.
Original entry on oeis.org
592417938, 806692194, 940415058, 980421939, 1269819378, 1355899923, 1488645939, 1599073938, 1635878754, 1657885698, 1666044963, 1758151458, 1797373314, 1813434483, 1991146899, 2064726483, 2198975058, 2246905683, 2266525314, 2302589298, 2302698258, 2502041283
Offset: 1
592417938 is a term because 592417938 = 6^4 + 59^4 + 65^4 + 154^4 = 7^4 + 11^4 + 20^4 + 156^4 = 10^4 + 17^4 + 17^4 + 156^4 = 12^4 + 112^4 + 115^4 + 127^4 = 15^4 + 86^4 + 107^4 + 142^4 = 21^4 + 49^4 + 70^4 + 154^4 = 25^4 + 107^4 + 112^4 + 132^4 = 26^4 + 45^4 + 71^4 + 154^4 = 28^4 + 105^4 + 112^4 + 133^4 = 63^4 + 77^4 + 112^4 + 140^4.
-
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, 4):
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])
A345188
Numbers that are the sum of five third powers in exactly ten ways.
Original entry on oeis.org
5860, 6588, 6651, 6859, 6947, 8056, 8289, 8569, 8758, 9045, 9099, 9227, 9414, 9612, 9829, 10009, 10277, 10485, 10522, 10529, 10800, 10963, 10970, 11008, 11061, 11089, 11241, 11385, 11458, 11656, 11719, 11782, 11817, 11845, 11934, 11990, 12016, 12060, 12088
Offset: 1
6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3 = 1^3 + 4^3 + 6^3 + 13^3 + 14^3 = 1^3 + 5^3 + 8^3 + 8^3 + 16^3 = 1^3 + 10^3 + 10^3 + 11^3 + 12^3 = 2^3 + 2^3 + 9^3 + 12^3 + 14^3 = 2^3 + 3^3 + 8^3 + 11^3 + 15^3 = 3^3 + 8^3 + 8^3 + 11^3 + 14^3 = 3^3 + 3^3 + 5^3 + 10^3 + 16^3 = 5^3 + 5^3 + 8^3 + 10^3 + 15^3 = 8^3 + 9^3 + 10^3 + 10^3 + 12^3.
-
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 == 10])
for x in range(len(rets)):
print(rets[x])
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