A345850
Numbers that are the sum of nine fourth powers in exactly eight ways.
Original entry on oeis.org
6804, 6869, 8019, 8084, 8324, 8499, 8564, 9044, 9124, 9219, 9234, 9284, 9364, 9429, 9474, 9494, 9604, 9669, 9749, 9779, 10148, 10259, 10293, 10339, 10388, 10453, 10514, 10579, 10628, 10644, 10754, 10789, 11029, 11059, 11189, 11204, 11299, 11363, 11364, 11379
Offset: 1
6869 is a term because 6869 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 9^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.
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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, 9):
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])
A345625
Numbers that are the sum of nine fifth powers in eight or more ways.
Original entry on oeis.org
1431398, 1431640, 1531397, 1952415, 1969221, 2247917, 2530399, 2596936, 2652563, 2652860, 2736790, 2851254, 2965588, 3088909, 3148674, 3273590, 3297416, 3329120, 3329362, 3332244, 3336895, 3345442, 3345653, 3353186, 3361614, 3362217, 3364738, 3378178, 3553641
Offset: 1
1431640 is a term because 1431640 = 1^5 + 2^5 + 3^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 3^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 1^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 9^5 + 12^5 + 16^5 = 2^5 + 2^5 + 3^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 2^5 + 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 3^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5.
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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, 9):
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])
A346333
Numbers that are the sum of eight fifth powers in exactly eight ways.
Original entry on oeis.org
8625619, 9773236, 10036233, 10071050, 12247994, 13180706, 13377868, 13662501, 14584992, 14591744, 14611077, 15251119, 16112362, 16374250, 16391025, 16472544, 16588000, 16667851, 17059075, 17216298, 17405300, 17917097, 18107564, 18392902, 18470839, 18541635
Offset: 1
8625619 is a term because 8625619 = 2^5 + 5^5 + 5^5 + 9^5 + 10^5 + 12^5 + 12^5 + 24^5 = 1^5 + 3^5 + 8^5 + 9^5 + 11^5 + 11^5 + 12^5 + 24^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 16^5 + 23^5 = 1^5 + 3^5 + 3^5 + 4^5 + 11^5 + 17^5 + 18^5 + 22^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 16^5 + 22^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 16^5 + 19^5 + 20^5 = 3^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 18^5 + 20^5 = 3^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 18^5 + 20^5.
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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, 8):
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])
A346342
Numbers that are the sum of nine fifth powers in exactly seven ways.
Original entry on oeis.org
1431429, 1439173, 1447570, 1504636, 1597929, 1671167, 1696159, 1697686, 1697928, 1778835, 1936454, 1975049, 2017344, 2092122, 2182161, 2198967, 2208680, 2280818, 2283911, 2289343, 2314335, 2329845, 2340319, 2345806, 2362370, 2388651, 2497771, 2529407, 2530672
Offset: 1
1431398 is a term because 1431398 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 12^5 + 16^5 = 1^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.
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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, 9):
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])
A346344
Numbers that are the sum of nine fifth powers in exactly nine ways.
Original entry on oeis.org
1969221, 2596936, 3353186, 3378178, 3923426, 3981447, 4094027, 4096729, 4112329, 4114188, 4129465, 4137209, 4147736, 4170112, 4172994, 4254304, 4303773, 4410482, 4475846, 4477936, 4483379, 4485480, 4501441, 4543232, 4652011, 4691855, 4724015, 4733970, 4750241
Offset: 1
1969221 is a term because 1969221 = 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 + 16^5 = 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 12^5 + 12^5 + 13^5 + 16^5 = 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 12^5 + 12^5 + 13^5 + 16^5 = 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 14^5 + 15^5 = 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 14^5 + 15^5 = 1^5 + 4^5 + 5^5 + 8^5 + 9^5 + 13^5 + 13^5 + 13^5 + 15^5 = 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 + 14^5 = 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 + 14^5.
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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, 9):
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])
A346353
Numbers that are the sum of ten fifth powers in exactly eight ways.
Original entry on oeis.org
944383, 953139, 953414, 985453, 1118585, 1151438, 1185375, 1198879, 1206546, 1209912, 1216569, 1217172, 1218912, 1223321, 1225398, 1234631, 1241834, 1251195, 1251406, 1252123, 1259685, 1265563, 1265594, 1267937, 1275375, 1281736, 1295418, 1297697, 1298088
Offset: 1
944383 is a term because 944383 = 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 + 10^5 + 12^5 + 14^5 = 2^5 + 4^5 + 5^5 + 5^5 + 7^5 + 7^5 + 7^5 + 10^5 + 12^5 + 14^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 1^5 + 3^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 12^5.
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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, 10):
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])
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