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 1-6 of 6 results.

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])

A345519 Numbers that are the sum of six cubes in ten or more ways.

Original entry on oeis.org

3104, 3215, 3222, 3267, 3313, 3339, 3374, 3430, 3465, 3491, 3493, 3528, 3547, 3584, 3645, 3654, 3698, 3708, 3736, 3745, 3752, 3754, 3761, 3771, 3780, 3789, 3817, 3824, 3843, 3862, 3869, 3878, 3880, 3888, 3906, 3915, 3923, 3943, 3950, 3969, 3995, 4004, 4014
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			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.

Links

Crossrefs

Cf. A345187, A345477, A345506, A345518, A345567, A345772.

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

A345540 Numbers that are the sum of eight cubes in ten or more ways.

Original entry on oeis.org

1185, 1243, 1262, 1288, 1295, 1299, 1386, 1393, 1397, 1400, 1412, 1419, 1423, 1448, 1449, 1451, 1458, 1460, 1464, 1467, 1475, 1477, 1497, 1501, 1503, 1504, 1505, 1512, 1513, 1514, 1516, 1521, 1523, 1539, 1540, 1541, 1542, 1553, 1558, 1559, 1560, 1565, 1566
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			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.

Links

Crossrefs

Cf. A345497, A345506, A345539, A345549, A345585, A345792.

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

A345576 Numbers that are the sum of seven fourth powers in ten or more ways.

Original entry on oeis.org

31251, 44547, 45827, 45892, 45907, 47667, 47971, 48292, 49572, 49812, 50052, 51092, 52372, 53316, 53476, 54531, 54596, 54756, 54996, 57411, 58036, 58116, 58276, 58516, 58660, 58756, 59331, 59781, 59796, 59811, 59827, 59861, 59876, 59892, 60036, 60371, 60436
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			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.

Links

Crossrefs

Cf. A345506, A345567, A345575, A345585, A345643, A345832.

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

A345782 Numbers that are the sum of seven cubes in exactly ten ways.

Original entry on oeis.org

1704, 1711, 1800, 1837, 1863, 1926, 1938, 1963, 2008, 2019, 2045, 2053, 2059, 2078, 2113, 2143, 2161, 2171, 2176, 2217, 2223, 2250, 2260, 2266, 2276, 2286, 2295, 2304, 2313, 2315, 2331, 2350, 2354, 2357, 2374, 2404, 2412, 2413, 2442, 2444, 2446, 2447, 2511
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345506 at term 3 because 1774 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 12^3 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3 + 6^3 + 11^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 9^3 + 10^3 = 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 11^3 = 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 + 9^3 = 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.
Likely finite.

Examples

			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.

Links

Crossrefs

Cf. A345506, A345772, A345781, A345792, A345832.

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

A345487 Numbers that are the sum of seven squares in ten or more ways.

Original entry on oeis.org

70, 79, 82, 85, 87, 88, 90, 91, 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, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			79 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 8^2
   = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 3^2 + 3^2 + 4^2 + 6^2
   = 2^2 + 3^2 + 3^2 + 3^2 + 4^2 + 4^2 + 4^2
   = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2
so 79 is a term.

Links

Crossrefs

Cf. A345477, A345486, A345497, A345506.

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

Formula

Conjectures from Chai Wah Wu, Jan 05 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 10.
G.f.: x*(-x^9 + x^8 - x^7 + x^6 - x^5 - x^4 - 6*x^2 - 61*x + 70)/(x - 1)^2. (End)
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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