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.

A345185 Numbers that are the sum of five third powers in nine or more ways.

Original entry on oeis.org

5860, 6112, 6138, 6462, 6497, 6588, 6651, 6859, 6947, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8056, 8163, 8289, 8352, 8371, 8443, 8506, 8560, 8569, 8630, 8632, 8758, 8928, 8991, 9017, 9045, 9080, 9099
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3  = 1^3 + 3^3 + 7^3 + 12^3 + 14^3  = 1^3 + 6^3 + 6^3 + 7^3 + 16^3  = 2^3 + 2^3 + 9^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 11^3 + 15^3  = 2^3 + 8^3 + 9^3 + 9^3 + 14^3  = 3^3 + 3^3 + 3^3 + 4^3 + 17^3  = 3^3 + 5^3 + 8^3 + 11^3 + 14^3  = 8^3 + 8^3 + 8^3 + 11^3 + 12^3.

Links

Crossrefs

Cf. A341891, A344802, A345146, A345183, A345186, A345187, A345518.

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

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])

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])

A345566 Numbers that are the sum of six fourth powers in nine or more ways.

Original entry on oeis.org

88595, 122915, 132546, 134931, 144835, 146450, 151556, 161475, 162355, 162755, 170275, 171555, 171795, 172036, 172835, 173075, 177380, 177716, 180770, 183540, 183620, 184835, 185315, 185555, 187700, 187715, 190100, 190211, 193635, 195380, 195780, 196435
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			122915 is a term because 122915 = 1^4 + 3^4 + 6^4 + 9^4 + 10^4 + 18^4 = 1^4 + 4^4 + 7^4 + 8^4 + 15^4 + 16^4 = 1^4 + 7^4 + 9^4 + 10^4 + 14^4 + 16^4 = 2^4 + 3^4 + 4^4 + 5^4 + 14^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 11^4 + 18^4 = 2^4 + 9^4 + 9^4 + 12^4 + 14^4 + 15^4 = 3^4 + 5^4 + 6^4 + 6^4 + 11^4 + 18^4 = 3^4 + 8^4 + 10^4 + 11^4 + 13^4 + 16^4 = 5^4 + 6^4 + 7^4 + 11^4 + 14^4 + 16^4 = 8^4 + 8^4 + 9^4 + 10^4 + 11^4 + 17^4.

Links

Crossrefs

Cf. A341891, A345518, A345565, A345567, A345575, A345723, A345821.

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 >= 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])

A345771 Numbers that are the sum of six cubes in exactly nine ways.

Original entry on oeis.org

2438, 2457, 2494, 2555, 2593, 2709, 2772, 2889, 2942, 2980, 3033, 3043, 3096, 3160, 3195, 3241, 3250, 3257, 3276, 3402, 3427, 3437, 3467, 3556, 3582, 3592, 3608, 3609, 3617, 3672, 3735, 3825, 3850, 3852, 3871, 3924, 3934, 3962, 3976, 3979, 3996, 3997, 4006
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345518 at term 14 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

			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. A345186, A345518, A345770, A345772, A345781, A345821.

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])

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)
Showing 1-7 of 7 results.

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

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