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 1-5 of 5 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])

A345476 Numbers that are the sum of six squares in nine or more ways.

Original entry on oeis.org

78, 81, 84, 86, 89, 92, 93, 95, 99, 100, 101, 102, 104, 105, 107, 108, 110, 111, 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, 140, 141, 142, 143, 144, 145, 146, 147, 148
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			81 = 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 5^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 5^2 + 6^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 8^2
   = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 7^2
   = 1^2 + 4^2 + 4^2 + 4^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 7^2
   = 2^2 + 2^2 + 4^2 + 4^2 + 4^2 + 5^2
   = 2^2 + 3^2 + 3^2 + 3^2 + 5^2 + 5^2
   = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 6^2
so 81 is a term.

Links

Crossrefs

Cf. A344802, A344812, A345477, A345486, A345518.

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, 6):
    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])

Formula

Conjectures from Chai Wah Wu, Jan 05 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 20.
G.f.: x*(-x^19 + x^18 - x^17 + x^16 - x^15 + x^14 - x^13 + x^12 - 3*x^9 + 2*x^8 + x^7 - 2*x^6 + x^4 - x^3 - 75*x + 78)/(x - 1)^2. (End)

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])

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)

A345496 Numbers that are the sum of eight squares in nine or more ways.

Original entry on oeis.org

62, 64, 67, 70, 71, 73, 74, 76, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88, 89, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			64 is a term because 64 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2.

Links

Crossrefs

Cf. A345486, A345495, A345497, A345539.

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, 8):
    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])
Showing 1-5 of 5 results.

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

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