A345153 -id:A345153 - cp's OEIS frontend

A345152 Numbers that are the sum of four third powers in eight or more ways.

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

Offset: 1

David Consiglio, Jr., Jun 09 2021

nonn

			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.

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Cf. A025373, A344924, A345087, A345146, A345150, A345153, A345183.

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])

A344925 Numbers that are the sum of four fourth powers in exactly eight ways.

Original entry on oeis.org

13155858, 26421474, 35965458, 39803778, 98926434, 128198994, 143776179, 156279618, 210493728, 237073554, 248075538, 255831858, 257931378, 269965938, 270289698, 292967619, 293579874, 295880274, 300120003, 301080243, 302115843, 305670834, 309742434, 331957458

Offset: 1

David Consiglio, Jr., Jun 02 2021

nonn

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.

			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.

David Consiglio, Jr., Table of n, a(n) for n = 1..100

Cf. A344738, A344923, A344924, A344927, A344945, A345153.

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])

A345088 Numbers that are the sum of three third powers in exactly eight ways.

Original entry on oeis.org

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

Offset: 1

David Consiglio, Jr., Jun 07 2021

nonn

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.

			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.

David Consiglio, Jr., Table of n, a(n) for n = 1..100

Cf. A025328, A344738, A345085, A345087, A345120, A345153.

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])

A345151 Numbers that are the sum of four third powers in exactly seven ways.

Original entry on oeis.org

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

Offset: 1

David Consiglio, Jr., Jun 09 2021

nonn

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.

			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.

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Cf. A025363, A344923, A345085, A345149, A345150, A345153, A345181.

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])

A345154 Numbers that are the sum of four third powers in exactly nine ways.

Original entry on oeis.org

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

Offset: 1

David Consiglio, Jr., Jun 09 2021

nonn

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.

			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.

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Cf. A025365, A344927, A345120, A345146, A345153, A345156, A345186.

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])

A345184 Numbers that are the sum of five third powers in exactly eight ways.

Original entry on oeis.org

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

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

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.

			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.

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Cf. A294742, A344945, A345153, A345181, A345183, A345186, A345770.

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])

cp's OEIS Frontend

A345152 Numbers that are the sum of four third powers in eight or more ways.

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A344925 Numbers that are the sum of four fourth powers in exactly eight ways.

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A345088 Numbers that are the sum of three third powers in exactly eight ways.

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A345151 Numbers that are the sum of four third powers in exactly seven ways.

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A345154 Numbers that are the sum of four third powers in exactly nine ways.

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A345184 Numbers that are the sum of five third powers in exactly eight ways.

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