A345184 -id:A345184 - cp's OEIS frontend

A345183 Numbers that are the sum of five third powers in eight or more ways.

4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5860, 5914, 6075, 6112, 6138, 6202, 6462, 6497, 6499, 6560, 6588, 6616, 6642, 6651, 6677, 6833, 6859, 6884, 6947, 7001, 7008, 7038, 7057, 7064, 7099, 7111, 7128, 7155, 7190, 7218, 7316

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

			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. A344801, A344944, A345152, A345180, A345184, A345185, A345517.

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])

A344945 Numbers that are the sum of five fourth powers in exactly eight ways.

Original entry on oeis.org

534130, 654754, 663155, 729219, 737459, 742770, 758354, 810034, 813459, 816579, 831250, 906034, 930499, 954930, 1009954, 1055619, 1083955, 1099459, 1101859, 1103554, 1106019, 1157634, 1167794, 1180003, 1215394, 1217539, 1246354, 1253074, 1255539, 1278690

Offset: 1

David Consiglio, Jr., Jun 03 2021

nonn

Differs from A344944 at term 2 because 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.

			534130 is a term because 534130 = 1^4 + 3^4 + 16^4 + 22^4 + 22^4  = 2^4 + 2^4 + 4^4 + 7^4 + 27^4  = 2^4 + 3^4 + 6^4 + 6^4 + 27^4  = 2^4 + 6^4 + 9^4 + 21^4 + 24^4  = 4^4 + 16^4 + 17^4 + 18^4 + 23^4  = 6^4 + 8^4 + 11^4 + 22^4 + 23^4  = 7^4 + 8^4 + 16^4 + 19^4 + 24^4  = 13^4 + 14^4 + 14^4 + 21^4 + 22^4.

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

Cf. A341892, A344925, A344943, A344944, A345184, A345820.

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 == 8])
for x in range(len(rets)):
    print(rets[x])

A345153 Numbers that are the sum of four third powers in exactly eight ways.

Original entry on oeis.org

27720, 30429, 31339, 31402, 33579, 34624, 34776, 36162, 40105, 42695, 44037, 44163, 44226, 44947, 45162, 45675, 46277, 46900, 47600, 49042, 50112, 50689, 51058, 51597, 51805, 52227, 52264, 52507, 53144, 54271, 54873, 55692, 55790, 56240, 58032, 58221, 58312

Offset: 1

David Consiglio, Jr., Jun 09 2021

nonn

Differs from A345152 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.

			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. A025364, A344925, A345088, A345151, A345152, A345154, A345184.

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])

A345181 Numbers that are the sum of five third powers in exactly seven ways.

Original entry on oeis.org

4472, 4544, 4600, 4957, 5076, 5113, 5120, 5132, 5165, 5174, 5347, 5354, 5384, 5391, 5410, 5445, 5474, 5481, 5507, 5543, 5617, 5715, 5760, 5834, 5895, 5923, 5984, 5986, 6049, 6128, 6131, 6245, 6280, 6373, 6407, 6434, 6436, 6544, 6553, 6733, 6768, 6831, 6840

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

Differs from A345180 at term 1 because 4392 = 1^3 + 1^3 + 10^3 + 10^3 + 11^3 = 1^3 + 2^3 + 2^3 + 9^3 + 14^3 = 1^3 + 8^3 + 9^3 + 10^3 + 10^3 = 2^3 + 2^3 + 3^3 + 5^3 + 15^3 = 2^3 + 3^3 + 5^3 + 8^3 + 14^3 = 2^3 + 8^3 + 8^3 + 8^3 + 12^3 = 3^3 + 6^3 + 7^3 + 8^3 + 13^3 = 5^3 + 5^3 + 5^3 + 9^3 + 13^3.

			4472 is a term because 4472 = 1^3 + 4^3 + 4^3 + 4^3 + 15^3  = 2^3 + 2^3 + 9^3 + 11^3 + 11^3  = 2^3 + 3^3 + 4^3 + 5^3 + 15^3  = 2^3 + 3^3 + 7^3 + 11^3 + 12^3  = 3^3 + 3^3 + 6^3 + 10^3 + 13^3  = 3^3 + 4^3 + 5^3 + 8^3 + 14^3  = 5^3 + 5^3 + 7^3 + 10^3 + 12^3.

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

Cf. A294741, A344943, A345151, A345175, A345180, A345184, A345769.

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 == 7])
for x in range(len(rets)):
    print(rets[x])

A345186 Numbers that are the sum of five third powers in exactly nine ways.

Original entry on oeis.org

6112, 6138, 6462, 6497, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8163, 8352, 8443, 8560, 8630, 8632, 8928, 8991, 9017, 9136, 9143, 9171, 9288, 9316, 9379, 9505, 9566, 9647, 9658, 9675, 9684, 9745, 9773

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

Differs from A345185 at term 1 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.

			6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3  = 1^3 + 3^3 + 7^3 + 12^3 + 14^3  = 1^3 + 6^3 + 6^3 + 7^3 + 16^3  = 2^3 + 2^3 + 9^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 11^3 + 15^3  = 2^3 + 8^3 + 9^3 + 9^3 + 14^3  = 3^3 + 3^3 + 3^3 + 4^3 + 17^3  = 3^3 + 5^3 + 8^3 + 11^3 + 14^3  = 8^3 + 8^3 + 8^3 + 11^3 + 12^3.

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

Cf. A294743, A341892, A345154, A345184, A345185, A345188, A345771.

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 == 9])
for x in range(len(rets)):
    print(rets[x])

A345770 Numbers that are the sum of six cubes in exactly eight ways.

Original entry on oeis.org

1981, 2105, 2168, 2277, 2368, 2376, 2431, 2466, 2538, 2557, 2583, 2646, 2665, 2672, 2746, 2753, 2763, 2765, 2880, 2881, 2916, 2961, 2970, 2977, 2979, 2987, 3007, 3040, 3042, 3049, 3068, 3088, 3141, 3159, 3169, 3185, 3248, 3278, 3311, 3312, 3367, 3384, 3393

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345517 at term 8 because 2438 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 12^3 = 1^3 + 5^3 + 5^3 + 9^3 + 9^3 + 9^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 2^3 + 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 2^3 + 4^3 + 5^3 + 8^3 + 9^3 + 10^3 = 4^3 + 4^3 + 5^3 + 5^3 + 9^3 + 11^3 = 6^3 + 7^3 + 7^3 + 8^3 + 8^3 + 8^3.

			2105 is a term because 2105 = 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 11^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 11^3 = 1^3 + 2^3 + 6^3 + 7^3 + 7^3 + 8^3 = 1^3 + 4^3 + 4^3 + 4^3 + 8^3 + 9^3 = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 10^3 = 2^3 + 3^3 + 4^3 + 5^3 + 8^3 + 9^3 = 3^3 + 3^3 + 3^3 + 7^3 + 7^3 + 9^3 = 5^3 + 5^3 + 5^3 + 5^3 + 7^3 + 8^3.

Sean A. Irvine, Table of n, a(n) for n = 1..1347

Cf. A345184, A345517, A345769, A345771, A345780, A345820.

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, 6):
    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

A345183 Numbers that are the sum of five third powers in eight or more ways.

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A344945 Numbers that are the sum of five fourth powers in exactly eight ways.

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A345153 Numbers that are the sum of four third powers in exactly eight ways.

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A345181 Numbers that are the sum of five third powers in exactly seven ways.

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A345186 Numbers that are the sum of five third powers in exactly nine ways.

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A345770 Numbers that are the sum of six cubes in exactly eight ways.

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