A345832
Numbers that are the sum of seven fourth powers in exactly ten ways.
Original entry on oeis.org
31251, 44547, 45827, 45892, 47667, 47971, 49572, 51092, 53316, 53476, 54531, 54596, 54756, 57411, 58276, 58660, 59781, 59811, 59827, 59861, 59876, 59892, 61076, 64581, 65876, 65891, 66356, 66596, 66676, 67716, 67876, 68131, 68322, 68772, 69171, 69667, 70116
Offset: 1
44547 is a term because 44547 = 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 11^4 + 13^4 = 1^4 + 2^4 + 2^4 + 6^4 + 7^4 + 7^4 + 14^4 = 1^4 + 2^4 + 6^4 + 6^4 + 9^4 + 11^4 + 12^4 = 1^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 13^4 = 2^4 + 2^4 + 8^4 + 9^4 + 9^4 + 9^4 + 12^4 = 2^4 + 4^4 + 6^4 + 6^4 + 9^4 + 9^4 + 13^4 = 2^4 + 4^4 + 7^4 + 7^4 + 8^4 + 11^4 + 12^4 = 3^4 + 3^4 + 4^4 + 4^4 + 7^4 + 12^4 + 12^4 = 3^4 + 6^4 + 6^4 + 7^4 + 8^4 + 11^4 + 12^4 = 4^4 + 4^4 + 8^4 + 8^4 + 9^4 + 11^4 + 11^4.
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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, 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])
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.
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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 == 9])
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])
A346335
Numbers that are the sum of eight fifth powers in exactly ten ways.
Original entry on oeis.org
15539667, 22932525, 24393600, 24650406, 24952961, 24953742, 25142513, 26001294, 27988486, 28609075, 29309819, 31794336, 32223105, 32527286, 32610600, 32807777, 32890541, 32998317, 33015125, 33187858, 33361339, 33550572, 33659175, 33782597, 34029369, 34073650
Offset: 1
15539667 is a term because 15539667 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 24^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 22^5 + 24^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 19^5 + 20^5 + 24^5 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 24^5 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5.
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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 == 10])
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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