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.
		

Crossrefs

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])

A345821 Numbers that are the sum of six fourth powers in exactly nine ways.

Original entry on oeis.org

88595, 132546, 134931, 144835, 146450, 162355, 170275, 171555, 171795, 172036, 172835, 177380, 177716, 180770, 183540, 184835, 185555, 187700, 187715, 190100, 190211, 193635, 195380, 195780, 196435, 197780, 199075, 199475, 199730, 199955, 202196, 202980
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345566 at term 2 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.

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.
		

Crossrefs

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])

A345186 Numbers that are the sum of five third powers in exactly nine ways.

Original entry on oeis.org

6112, 6138, 6462, 6497, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8163, 8352, 8443, 8560, 8630, 8632, 8928, 8991, 9017, 9136, 9143, 9171, 9288, 9316, 9379, 9505, 9566, 9647, 9658, 9675, 9684, 9745, 9773
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Comments

Differs from A345185 at term 1 because 5860 = 1^3 + 1^3 + 5^3 + 8^3 + 16^3 = 1^3 + 2^3 + 3^3 + 11^3 + 15^3 = 1^3 + 3^3 + 8^3 + 11^3 + 14^3 = 1^3 + 5^3 + 5^3 + 10^3 + 15^3 = 1^3 + 9^3 + 10^3 + 10^3 + 12^3 = 2^3 + 3^3 + 8^3 + 9^3 + 15^3 = 2^3 + 3^3 + 5^3 + 12^3 + 14^3 = 2^3 + 8^3 + 8^3 + 12^3 + 12^3 = 3^3 + 8^3 + 8^3 + 9^3 + 14^3 = 3^3 + 6^3 + 7^3 + 12^3 + 13^3.

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.
		

Crossrefs

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])

A345770 Numbers that are the sum of six cubes in exactly eight ways.

Original entry on oeis.org

1981, 2105, 2168, 2277, 2368, 2376, 2431, 2466, 2538, 2557, 2583, 2646, 2665, 2672, 2746, 2753, 2763, 2765, 2880, 2881, 2916, 2961, 2970, 2977, 2979, 2987, 3007, 3040, 3042, 3049, 3068, 3088, 3141, 3159, 3169, 3185, 3248, 3278, 3311, 3312, 3367, 3384, 3393
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345517 at term 8 because 2438 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 12^3 = 1^3 + 5^3 + 5^3 + 9^3 + 9^3 + 9^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 2^3 + 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 2^3 + 4^3 + 5^3 + 8^3 + 9^3 + 10^3 = 4^3 + 4^3 + 5^3 + 5^3 + 9^3 + 11^3 = 6^3 + 7^3 + 7^3 + 8^3 + 8^3 + 8^3.

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.
		

Crossrefs

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])

A345781 Numbers that are the sum of seven cubes in exactly nine ways.

Original entry on oeis.org

1496, 1648, 1720, 1737, 1772, 1781, 1802, 1835, 1844, 1882, 1891, 1898, 1900, 1907, 1912, 1919, 1945, 1952, 1954, 1961, 1996, 2000, 2012, 2026, 2071, 2080, 2098, 2107, 2110, 2115, 2116, 2132, 2134, 2136, 2139, 2150, 2152, 2168, 2178, 2185, 2187, 2195, 2205
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345527 at term 3 because 1704 = 1^3 + 1^3 + 1^3 + 3^3 + 6^3 + 9^3 + 9^3 = 1^3 + 1^3 + 1^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 1^3 + 2^3 + 2^3 + 7^3 + 7^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 6^3 + 11^3 = 1^3 + 2^3 + 4^3 + 6^3 + 7^3 + 7^3 + 9^3 = 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 6^3 + 11^3 = 2^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 7^3 + 10^3 = 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 + 9^3 = 3^3 + 6^3 + 6^3 + 6^3 + 7^3 + 7^3 + 7^3.
Likely finite.

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.
		

Crossrefs

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])

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.
		

Crossrefs

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])
Showing 1-6 of 6 results.