cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A345517 Numbers that are the sum of six cubes in eight or more ways.

Original entry on oeis.org

1981, 2105, 2168, 2277, 2368, 2376, 2431, 2438, 2457, 2466, 2494, 2538, 2555, 2557, 2583, 2593, 2646, 2665, 2672, 2709, 2746, 2753, 2763, 2765, 2772, 2880, 2881, 2889, 2916, 2942, 2961, 2970, 2977, 2979, 2980, 2987, 3007, 3033, 3040, 3042, 3043, 3049, 3068
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			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.

Links

Crossrefs

Cf. A344812, A345183, A345516, A345518, A345526, A345565, A345770.

Programs

  • Python
    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])

A345525 Numbers that are the sum of seven cubes in seven or more ways.

Original entry on oeis.org

1072, 1170, 1235, 1261, 1268, 1305, 1385, 1392, 1396, 1411, 1440, 1441, 1448, 1450, 1459, 1489, 1496, 1502, 1504, 1513, 1515, 1538, 1540, 1547, 1552, 1557, 1559, 1564, 1565, 1566, 1567, 1576, 1585, 1587, 1592, 1593, 1594, 1600, 1602, 1603, 1606, 1613, 1620
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1170 is a term because 1170 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3 = 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 7^3.

Links

Crossrefs

Cf. A345484, A345516, A345524, A345526, A345537, A345573, A345779.

Programs

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

A345527 Numbers that are the sum of seven cubes in nine or more ways.

Original entry on oeis.org

1496, 1648, 1704, 1711, 1720, 1737, 1772, 1774, 1781, 1800, 1802, 1835, 1837, 1844, 1863, 1882, 1889, 1891, 1893, 1898, 1900, 1907, 1912, 1919, 1926, 1938, 1945, 1952, 1954, 1961, 1963, 1982, 1989, 1996, 2000, 2008, 2012, 2015, 2019, 2026, 2045, 2052, 2053
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			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.

Links

Crossrefs

Cf. A345486, A345506, A345518, A345526, A345539, A345575, A345781.

Programs

  • Python
    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])

A345538 Numbers that are the sum of eight cubes in eight or more ways.

Original entry on oeis.org

970, 977, 984, 1054, 1073, 1075, 1080, 1090, 1099, 1106, 1110, 1125, 1129, 1136, 1148, 1160, 1166, 1171, 1178, 1181, 1185, 1186, 1188, 1192, 1197, 1204, 1206, 1211, 1217, 1218, 1223, 1225, 1230, 1232, 1234, 1236, 1237, 1242, 1243, 1249, 1262, 1263, 1269, 1273
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			977 is a term because 977 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 8^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A345495, A345526, A345537, A345539, A345547, A345583, A345790.

Programs

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

A345574 Numbers that are the sum of seven fourth powers in eight or more ways.

Original entry on oeis.org

19491, 21252, 21267, 21332, 21507, 21636, 21876, 23652, 25347, 30372, 31251, 31412, 31652, 32116, 32356, 33811, 33907, 35427, 35637, 35652, 35892, 36052, 36261, 37812, 37827, 38052, 38067, 38596, 38676, 39267, 39347, 39891, 39971, 39972, 40212, 40356, 40452
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			21252 is a term because 21252 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 1^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 7^4 + 8^4 + 9^4 + 9^4.

Links

Crossrefs

Cf. A345526, A345565, A345573, A345575, A345583, A345630, A345830.

Programs

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

A345780 Numbers that are the sum of seven cubes in exactly eight ways.

Original entry on oeis.org

1385, 1515, 1552, 1557, 1585, 1587, 1603, 1613, 1622, 1655, 1665, 1674, 1681, 1718, 1719, 1739, 1741, 1746, 1753, 1755, 1765, 1767, 1782, 1793, 1805, 1809, 1811, 1818, 1819, 1826, 1828, 1830, 1833, 1838, 1856, 1870, 1873, 1881, 1901, 1905, 1931, 1935, 1937
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345526 at term 2 because 1496 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 9^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 11^3 = 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 8^3 + 9^3 = 1^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 + 9^3 = 1^3 + 5^3 + 5^3 + 6^3 + 7^3 + 7^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 11^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 + 10^3 = 2^3 + 3^3 + 6^3 + 6^3 + 7^3 + 7^3 + 7^3 = 4^3 + 4^3 + 4^3 + 4^3 + 6^3 + 8^3 + 8^3.
Likely finite.

Examples

			1496 is a term because 1496 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 2^3 + 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3.

Links

Crossrefs

Cf. A345526, A345770, A345779, A345781, A345790, A345830.

Programs

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

A345485 Numbers that are the sum of seven squares in eight or more ways.

Original entry on oeis.org

61, 66, 69, 70, 72, 73, 76, 77, 78, 79, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			66 is a term because 66 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 7^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 6^2 = 1^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2.

Links

Crossrefs

Cf. A344812, A345484, A345486, A345495, A345526.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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 >= 8])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-7 of 7 results.

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