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.

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])

A345187 Numbers that are the sum of five third powers in ten or more ways.

Original entry on oeis.org

5860, 6588, 6651, 6859, 6947, 8056, 8289, 8371, 8506, 8569, 8758, 9045, 9080, 9099, 9108, 9227, 9414, 9612, 9801, 9829, 9864, 10009, 10018, 10044, 10277, 10466, 10485, 10522, 10529, 10800, 10963, 10970, 10979, 11008, 11017, 11061, 11089, 11152, 11241, 11385
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3  = 1^3 + 4^3 + 6^3 + 13^3 + 14^3  = 1^3 + 5^3 + 8^3 + 8^3 + 16^3  = 1^3 + 10^3 + 10^3 + 11^3 + 12^3  = 2^3 + 2^3 + 9^3 + 12^3 + 14^3  = 2^3 + 3^3 + 8^3 + 11^3 + 15^3  = 3^3 + 8^3 + 8^3 + 11^3 + 14^3  = 3^3 + 3^3 + 5^3 + 10^3 + 16^3  = 5^3 + 5^3 + 8^3 + 10^3 + 15^3  = 8^3 + 9^3 + 10^3 + 10^3 + 12^3.

Links

Crossrefs

Cf. A341897, A344803, A345155, A345185, A345188, A345519.

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, 5):
    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])

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

Original entry on oeis.org

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

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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

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])

A345567 Numbers that are the sum of six fourth powers in ten or more ways.

Original entry on oeis.org

122915, 151556, 161475, 162755, 173075, 183620, 185315, 197795, 199106, 199940, 201875, 201955, 202275, 204275, 204340, 204595, 206115, 207395, 209795, 211075, 212420, 213731, 217620, 217826, 217891, 218515, 221250, 223715, 223955, 224180, 224451, 225875
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			151556 is a term because 151556 = 1^4 + 2^4 + 2^4 + 9^4 + 11^4 + 19^4 = 1^4 + 2^4 + 3^4 + 7^4 + 16^4 + 17^4 = 1^4 + 8^4 + 11^4 + 12^4 + 13^4 + 17^4 = 2^4 + 3^4 + 7^4 + 8^4 + 11^4 + 19^4 = 3^4 + 3^4 + 3^4 + 4^4 + 12^4 + 19^4 = 3^4 + 4^4 + 11^4 + 11^4 + 14^4 + 17^4 = 3^4 + 4^4 + 13^4 + 13^4 + 13^4 + 16^4 = 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 19^4 = 4^4 + 7^4 + 11^4 + 11^4 + 11^4 + 18^4 = 4^4 + 8^4 + 9^4 + 13^4 + 13^4 + 17^4.

Links

Crossrefs

Cf. A341897, A344196, A345519, A345566, A345576, A345822.

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, 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])

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

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

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.

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. A345188, A345519, A345771, A345782, A345822.

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])

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 1-6 of 6 results.

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