A345567
Numbers that are the sum of six fourth powers in ten or more ways.
Original entry on oeis.org
122915, 151556, 161475, 162755, 173075, 183620, 185315, 197795, 199106, 199940, 201875, 201955, 202275, 204275, 204340, 204595, 206115, 207395, 209795, 211075, 212420, 213731, 217620, 217826, 217891, 218515, 221250, 223715, 223955, 224180, 224451, 225875
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.
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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 >= 10])
for x in range(len(rets)):
print(rets[x])
A345643
Numbers that are the sum of seven fifth powers in ten or more ways.
Original entry on oeis.org
134581976, 189642309, 219063107, 235438301, 252277376, 275782407, 281935070, 290928076, 300919884, 308188849, 309631268, 315635200, 322947868, 327287951, 335530174, 342030094, 358852218, 361946949, 379913293, 384699424, 387538625, 391133568
Offset: 1
189642309 is a term because 189642309 = 1^5 + 1^5 + 2^5 + 19^5 + 30^5 + 36^5 + 40^5 = 1^5 + 2^5 + 6^5 + 7^5 + 18^5 + 20^5 + 45^5 = 1^5 + 6^5 + 21^5 + 27^5 + 29^5 + 36^5 + 39^5 = 2^5 + 9^5 + 19^5 + 23^5 + 33^5 + 33^5 + 40^5 = 3^5 + 4^5 + 21^5 + 28^5 + 29^5 + 34^5 + 40^5 = 6^5 + 7^5 + 11^5 + 29^5 + 33^5 + 36^5 + 37^5 = 7^5 + 12^5 + 17^5 + 20^5 + 29^5 + 32^5 + 42^5 = 8^5 + 11^5 + 21^5 + 21^5 + 22^5 + 34^5 + 42^5 = 13^5 + 14^5 + 14^5 + 19^5 + 21^5 + 38^5 + 40^5 = 20^5 + 21^5 + 24^5 + 24^5 + 24^5 + 38^5 + 38^5.
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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 >= 10])
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.
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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])
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.
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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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