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.

A345174 Numbers that are the sum of five third powers in six or more ways.

Original entry on oeis.org

2430, 2979, 3214, 3249, 3312, 3492, 3520, 3737, 3753, 3788, 3816, 3842, 3942, 3968, 4121, 4185, 4213, 4267, 4355, 4392, 4411, 4418, 4446, 4453, 4456, 4465, 4472, 4482, 4509, 4544, 4563, 4600, 4626, 4663, 4670, 4723, 4753, 4896, 4905, 4915, 4924, 4938, 4941
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			2430 is a term because 2430 = 1^3 + 2^3 + 2^3 + 5^3 + 12^3  = 1^3 + 3^3 + 4^3 + 7^3 + 11^3  = 2^3 + 2^3 + 6^3 + 6^3 + 11^3  = 2^3 + 3^3 + 3^3 + 9^3 + 10^3  = 3^3 + 5^3 + 8^3 + 8^3 + 8^3  = 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, 5):
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 6])
    for x in range(len(rets)):
        print(rets[x])

A343988 Numbers that are the sum of five positive cubes in exactly five ways.

Original entry on oeis.org

1765, 1980, 2043, 2104, 2195, 2250, 2449, 2486, 2491, 2493, 2547, 2584, 2592, 2738, 2745, 2764, 2817, 2888, 2915, 2953, 2969, 3095, 3096, 3133, 3142, 3186, 3188, 3240, 3275, 3277, 3310, 3366, 3403, 3422, 3459, 3464, 3466, 3483, 3529, 3583, 3608, 3627, 3653, 3664, 3671, 3690, 3697, 3707, 3725, 3744, 3746, 3781
Offset: 1

Views

Author

David Consiglio, Jr., May 06 2021

Keywords

Comments

Differs from A343989 at term 7 because 2430 = 1^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 4^3 + 5^3 + 8^3 + 12^3 = 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 4^3 + 5^3 + 8^3 + 9^3 + 10^3.

Examples

			2043 is a term because 2043 = 1^3 + 4^3 + 5^3 + 5^3 + 12^3 = 2^3 + 2^3 + 3^3 + 10^3 + 10^3 = 2^3 + 3^3 + 4^3 + 6^3 + 12^3 = 4^3 + 5^3 + 5^3 + 9^3 + 10^3 = 4^3 + 6^3 + 6^3 + 6^3 + 11^3.
		

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(1,50)]
    for pos in cwr(power_terms,5):
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k,v in keep.items() if v == 5])
    for x in range(len(rets)):
        print(rets[x])

A344941 Numbers that are the sum of five fourth powers in exactly six ways.

Original entry on oeis.org

151300, 225890, 236194, 243235, 246674, 250834, 286114, 288579, 300835, 302130, 302210, 303235, 309059, 317795, 320195, 334819, 334899, 335443, 336210, 338914, 346835, 356899, 363379, 366995, 373234, 375619, 389875, 391154, 392259, 393314, 394354, 412339
Offset: 1

Views

Author

David Consiglio, Jr., Jun 03 2021

Keywords

Comments

Differs from A344940 at term 2 because 197779 = 1^4 + 5^4 + 6^4 + 16^4 + 19^4 = 1^4 + 7^4 + 11^4 + 12^4 + 20^4 = 1^4 + 10^4 + 12^4 + 17^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 21^4 = 3^4 + 5^4 + 6^4 + 6^4 + 21^4 = 4^4 + 7^4 + 9^4 + 13^4 + 20^4 = 11^4 + 13^4 + 14^4 + 15^4 + 16^4.

Examples

			151300 is a term because 151300 = 3^4 + 3^4 + 3^4 + 12^4 + 19^4  = 3^4 + 11^4 + 11^4 + 14^4 + 17^4  = 3^4 + 13^4 + 13^4 + 13^4 + 16^4  = 6^4 + 9^4 + 9^4 + 9^4 + 19^4  = 7^4 + 11^4 + 11^4 + 11^4 + 18^4  = 8^4 + 9^4 + 13^4 + 13^4 + 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, 5):
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 6])
    for x in range(len(rets)):
        print(rets[x])

A345149 Numbers that are the sum of four third powers in exactly six ways.

Original entry on oeis.org

6883, 12411, 13923, 14112, 14581, 14896, 14904, 15561, 15876, 16317, 16640, 17208, 17479, 17992, 18739, 18865, 19035, 19080, 19665, 19712, 19763, 19880, 20007, 20384, 20979, 21231, 21420, 21707, 22409, 22617, 23149, 23940, 24355, 25515, 25984, 26208, 26334
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Comments

Differs from A345148 at term 3 because 13104 = 1^3 + 10^3 + 16^3 + 18^3 = 1^3 + 11^3 + 14^3 + 19^3 = 2^3 + 9^3 + 15^3 + 19^3 = 4^3 + 6^3 + 14^3 + 20^3 = 4^3 + 9^3 + 10^3 + 21^3 = 5^3 + 7^3 + 11^3 + 21^3 = 8^3 + 9^3 + 14^3 + 19^3.

Examples

			6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3  = 2^3 + 4^3 + 14^3 + 14^3  = 3^3 + 7^3 + 7^3 + 17^3  = 3^3 + 10^3 + 13^3 + 13^3  = 4^3 + 10^3 + 10^3 + 15^3  = 7^3 + 8^3 + 8^3 + 16^3.
		

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, 4):
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 6])
    for x in range(len(rets)):
        print(rets[x])

A345181 Numbers that are the sum of five third powers in exactly seven ways.

Original entry on oeis.org

4472, 4544, 4600, 4957, 5076, 5113, 5120, 5132, 5165, 5174, 5347, 5354, 5384, 5391, 5410, 5445, 5474, 5481, 5507, 5543, 5617, 5715, 5760, 5834, 5895, 5923, 5984, 5986, 6049, 6128, 6131, 6245, 6280, 6373, 6407, 6434, 6436, 6544, 6553, 6733, 6768, 6831, 6840
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Comments

Differs from A345180 at term 1 because 4392 = 1^3 + 1^3 + 10^3 + 10^3 + 11^3 = 1^3 + 2^3 + 2^3 + 9^3 + 14^3 = 1^3 + 8^3 + 9^3 + 10^3 + 10^3 = 2^3 + 2^3 + 3^3 + 5^3 + 15^3 = 2^3 + 3^3 + 5^3 + 8^3 + 14^3 = 2^3 + 8^3 + 8^3 + 8^3 + 12^3 = 3^3 + 6^3 + 7^3 + 8^3 + 13^3 = 5^3 + 5^3 + 5^3 + 9^3 + 13^3.

Examples

			4472 is a term because 4472 = 1^3 + 4^3 + 4^3 + 4^3 + 15^3  = 2^3 + 2^3 + 9^3 + 11^3 + 11^3  = 2^3 + 3^3 + 4^3 + 5^3 + 15^3  = 2^3 + 3^3 + 7^3 + 11^3 + 12^3  = 3^3 + 3^3 + 6^3 + 10^3 + 13^3  = 3^3 + 4^3 + 5^3 + 8^3 + 14^3  = 5^3 + 5^3 + 7^3 + 10^3 + 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 == 7])
    for x in range(len(rets)):
        print(rets[x])

A345768 Numbers that are the sum of six cubes in exactly six ways.

Original entry on oeis.org

1377, 1488, 1586, 1595, 1647, 1673, 1677, 1738, 1764, 1799, 1829, 1836, 1837, 1862, 1881, 1890, 1911, 1953, 1955, 2007, 2011, 2014, 2018, 2025, 2044, 2070, 2079, 2097, 2107, 2108, 2142, 2153, 2170, 2177, 2203, 2214, 2216, 2222, 2223, 2226, 2229, 2252, 2258
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345515 at term 8 because 1710 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 6^3 + 11^3 = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.

Examples

			1488 is a term because 1488 = 1^3 + 1^3 + 1^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.
		

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 == 6])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.