A345782 -id:A345782 - cp's OEIS frontend

A345506 Numbers that are the sum of seven cubes in ten or more ways.

1704, 1711, 1774, 1800, 1837, 1863, 1889, 1893, 1926, 1938, 1963, 1982, 1989, 2008, 2015, 2019, 2045, 2052, 2053, 2059, 2078, 2097, 2106, 2113, 2143, 2161, 2169, 2171, 2176, 2197, 2204, 2217, 2223, 2224, 2227, 2230, 2234, 2241, 2250, 2260, 2266, 2267, 2276

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			1711 is a term because 1711 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 9^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 4^3 + 4^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345487, A345519, A345527, A345540, A345576, A345782.

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, 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])

A345781 Numbers that are the sum of seven cubes in exactly nine ways.

Original entry on oeis.org

1496, 1648, 1720, 1737, 1772, 1781, 1802, 1835, 1844, 1882, 1891, 1898, 1900, 1907, 1912, 1919, 1945, 1952, 1954, 1961, 1996, 2000, 2012, 2026, 2071, 2080, 2098, 2107, 2110, 2115, 2116, 2132, 2134, 2136, 2139, 2150, 2152, 2168, 2178, 2185, 2187, 2195, 2205

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345527 at term 3 because 1704 = 1^3 + 1^3 + 1^3 + 3^3 + 6^3 + 9^3 + 9^3 = 1^3 + 1^3 + 1^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 1^3 + 2^3 + 2^3 + 7^3 + 7^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 6^3 + 11^3 = 1^3 + 2^3 + 4^3 + 6^3 + 7^3 + 7^3 + 9^3 = 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 6^3 + 11^3 = 2^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 7^3 + 10^3 = 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 + 9^3 = 3^3 + 6^3 + 6^3 + 6^3 + 7^3 + 7^3 + 7^3.

Likely finite.

			1648 is a term because 1648 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 5^3 + 6^3 + 6^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 + 8^3 = 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 7^3 + 7^3.

Sean A. Irvine, Table of n, a(n) for n = 1..338

Cf. A345527, A345771, A345780, A345782, A345791, A345831.

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, 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])

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

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345576 at term 5 because 45907 = 1^4 + 1^4 + 3^4 + 4^4 + 8^4 + 12^4 + 12^4 = 1^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 + 13^4 = 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 11^4 + 13^4 = 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 11^4 + 13^4 = 2^4 + 2^4 + 4^4 + 7^4 + 7^4 + 7^4 + 14^4 = 2^4 + 3^4 + 6^4 + 6^4 + 7^4 + 7^4 + 14^4 = 2^4 + 4^4 + 6^4 + 7^4 + 9^4 + 11^4 + 12^4 = 2^4 + 5^4 + 5^4 + 10^4 + 10^4 + 10^4 + 11^4 = 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 9^4 + 14^4 = 3^4 + 6^4 + 6^4 + 6^4 + 9^4 + 11^4 + 12^4 = 4^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 + 13^4.

			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.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345576, A345782, A345822, A345831, A345842, A346259.

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])

A345772 Numbers that are the sum of six cubes in exactly ten ways.

Original entry on oeis.org

3215, 3267, 3313, 3339, 3374, 3465, 3493, 3528, 3547, 3584, 3645, 3654, 3698, 3736, 3745, 3752, 3754, 3780, 3789, 3843, 3869, 3878, 3880, 3888, 3906, 3915, 3923, 3950, 3995, 4004, 4014, 4041, 4067, 4122, 4148, 4211, 4212, 4214, 4265, 4266, 4268, 4338, 4349

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345519 at term 1 because 3104 = 1^3 + 2^3 + 7^3 + 8^3 + 8^3 + 12^3 = 1^3 + 5^3 + 5^3 + 5^3 + 10^3 + 12^3 = 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 14^3 = 2^3 + 3^3 + 4^3 + 7^3 + 11^3 + 11^3 = 2^3 + 3^3 + 5^3 + 6^3 + 10^3 + 12^3 = 2^3 + 7^3 + 8^3 + 8^3 + 9^3 + 10^3 = 3^3 + 3^3 + 5^3 + 6^3 + 8^3 + 13^3 = 4^3 + 5^3 + 7^3 + 8^3 + 9^3 + 11^3 = 5^3 + 5^3 + 5^3 + 9^3 + 10^3 + 10^3 = 5^3 + 6^3 + 6^3 + 6^3 + 10^3 + 11^3 = 6^3 + 6^3 + 6^3 + 6^3 + 8^3 + 12^3.

			3215 is a term because 3215 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 13^3 = 1^3 + 1^3 + 2^3 + 2^3 + 9^3 + 12^3 = 1^3 + 1^3 + 3^3 + 8^3 + 8^3 + 11^3 = 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 12^3 = 1^3 + 3^3 + 3^3 + 3^3 + 10^3 + 11^3 = 1^3 + 3^3 + 8^3 + 8^3 + 8^3 + 9^3 = 2^3 + 3^3 + 6^3 + 6^3 + 8^3 + 11^3 = 3^3 + 3^3 + 3^3 + 8^3 + 9^3 + 10^3 = 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 12^3 = 6^3 + 6^3 + 6^3 + 6^3 + 7^3 + 10^3.

Sean A. Irvine, Table of n, a(n) for n = 1..1454

Cf. A345188, A345519, A345771, A345782, A345822.

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 == 10])
    for x in range(len(rets)):
        print(rets[x])

A345792 Numbers that are the sum of eight cubes in exactly ten ways.

Original entry on oeis.org

1185, 1243, 1288, 1295, 1299, 1386, 1397, 1400, 1412, 1423, 1448, 1449, 1451, 1458, 1460, 1464, 1467, 1475, 1477, 1501, 1503, 1505, 1512, 1513, 1516, 1539, 1540, 1541, 1553, 1558, 1559, 1568, 1577, 1578, 1586, 1588, 1591, 1592, 1594, 1595, 1596, 1600, 1608

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345540 at term 3 because 1262 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 6^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 7^3 + 7^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 6^3 + 9^3 = 1^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 6^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 10^3 = 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 9^3.

Likely finite.

			1243 is a term because 1243 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 8^3 = 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 7^3.

Sean A. Irvine, Table of n, a(n) for n = 1..161

Cf. A345540, A345782, A345791, A345802, A345842.

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, 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])

cp's OEIS Frontend

A345506 Numbers that are the sum of seven cubes in ten or more ways.

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A345781 Numbers that are the sum of seven cubes in exactly nine ways.

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A345832 Numbers that are the sum of seven fourth powers in exactly ten ways.

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A345772 Numbers that are the sum of six cubes in exactly ten ways.

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A345792 Numbers that are the sum of eight cubes in exactly ten ways.

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