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.

A345518 Numbers that are the sum of six cubes in nine or more ways.

Original entry on oeis.org

2438, 2457, 2494, 2555, 2593, 2709, 2772, 2889, 2942, 2980, 3033, 3043, 3096, 3104, 3160, 3195, 3215, 3222, 3241, 3250, 3257, 3267, 3276, 3313, 3339, 3374, 3402, 3427, 3430, 3437, 3465, 3467, 3491, 3493, 3528, 3547, 3556, 3582, 3584, 3592, 3608, 3609, 3617
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2457 is a term because 2457 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 12^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 12^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 11^3 = 1^3 + 5^3 + 5^3 + 7^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 11^3 = 2^3 + 3^3 + 3^3 + 3^3 + 9^3 + 10^3 = 2^3 + 5^3 + 5^3 + 6^3 + 6^3 + 10^3 = 3^3 + 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 3^3 + 4^3 + 7^3 + 8^3 + 9^3.

Links

Crossrefs

Cf. A345185, A345476, A345517, A345519, A345527, A345566, A345771.

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

A344802 Numbers that are the sum of five squares in nine or more ways.

Original entry on oeis.org

101, 107, 109, 112, 115, 116, 118, 125, 127, 128, 131, 133, 134, 136, 139, 140, 142, 144, 146, 147, 148, 149, 151, 152, 154, 155, 157, 158, 159, 160, 161, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 178, 179, 180, 181, 182, 183, 184
Offset: 1

Views

Author

Sean A. Irvine, May 29 2021

Keywords

Links

Crossrefs

Cf. A025374, A294743, A344801, A344803, A345185, A345476.

A344812 Numbers that are the sum of six squares in eight or more ways.

Original entry on oeis.org

78, 81, 84, 86, 87, 89, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 107, 108, 109, 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
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. A344801, A344811, A345476, A345485, A345517.

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 >= 8])
    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 > 27.
G.f.: x*(-x^26 + x^25 - x^21 + x^20 - 2*x^7 + x^6 + x^5 - x^4 - x^3 - 75*x + 78)/(x - 1)^2. (End)

A345486 Numbers that are the sum of seven squares in nine or more ways.

Original entry on oeis.org

69, 70, 78, 79, 81, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			70 is a term because 70 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 8^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 5^2 + 5^2 = 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 + 5^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 7^2 = 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 6^2 = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2.

Links

Crossrefs

Cf. A345476, A345485, A345487, A345496, A345527.

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 >= 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 > 13.
G.f.: x*(-x^12 + x^11 - x^10 + x^9 - x^8 - x^7 + 2*x^6 - x^5 + x^4 - 7*x^3 + 7*x^2 - 68*x + 69)/(x - 1)^2. (End)

A345477 Numbers that are the sum of six squares in ten or more ways.

Original entry on oeis.org

81, 84, 86, 89, 92, 93, 95, 100, 101, 102, 104, 105, 107, 108, 110, 111, 113, 114, 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, 149, 150
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			84 = 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 8^2
   = 1^2 + 1^2 + 1^2 + 3^2 + 6^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 4^2 + 4^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 5^2 + 7^2
   = 1^2 + 1^2 + 4^2 + 4^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 5^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 3^2 + 3^2 + 5^2 + 6^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 8^2
   = 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 7^2
   = 2^2 + 4^2 + 4^2 + 4^2 + 4^2 + 4^2
   = 3^2 + 3^2 + 3^2 + 4^2 + 4^2 + 5^2
so 84 is a term.

Links

Crossrefs

Cf. A025430, A344803, A345476, A345487, A345519.

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 >= 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 > 20.
G.f.: x*(-x^19 + x^18 - x^17 + x^16 - x^15 + x^14 - x^13 + x^12 - x^11 + x^10 - 4*x^8 + 3*x^7 + x^6 - 2*x^5 + x^3 - x^2 - 78*x + 81)/(x - 1)^2. (End)
Showing 1-5 of 5 results.

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