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.

A345594 Numbers that are the sum of nine fourth powers in ten or more ways.

Original entry on oeis.org

9299, 12708, 12948, 13269, 13349, 13524, 13589, 13764, 13829, 13893, 14133, 14228, 14468, 14564, 14804, 14869, 14934, 14964, 15014, 15044, 15094, 15109, 15174, 15189, 15333, 15413, 15428, 15429, 15443, 15508, 15524, 15573, 15588, 15604, 15653, 15669, 15683
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			12708 is a term because 12708 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 10^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 10^4 = 1^4 + 1^4 + 1^4 + 5^4 + 6^4 + 6^4 + 6^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 8^4 + 9^4 = 1^4 + 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 6^4 + 7^4 + 9^4 = 1^4 + 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 10^4 = 2^4 + 2^4 + 3^4 + 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 = 2^4 + 4^4 + 4^4 + 4^4 + 5^4 + 7^4 + 7^4 + 7^4 + 8^4 = 3^4 + 4^4 + 4^4 + 5^4 + 6^4 + 6^4 + 7^4 + 7^4 + 8^4.
		

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, 9):
        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])

A345602 Numbers that are the sum of ten fourth powers in nine or more ways.

Original entry on oeis.org

6820, 6870, 6885, 6950, 7060, 7110, 7125, 7285, 7350, 7860, 7925, 7990, 8020, 8035, 8100, 8165, 8230, 8245, 8260, 8275, 8325, 8340, 8390, 8405, 8515, 8565, 8580, 8645, 8755, 8820, 8884, 8965, 8995, 9030, 9045, 9060, 9075, 9125, 9140, 9205, 9220, 9235, 9270
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6870 is a term because 6870 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 9^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4 + 7^4.
		

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, 10):
        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])

A345558 Numbers that are the sum of ten cubes in ten or more ways.

Original entry on oeis.org

721, 754, 756, 771, 782, 792, 797, 806, 808, 819, 832, 834, 845, 847, 848, 850, 860, 862, 867, 869, 871, 874, 876, 877, 878, 881, 884, 886, 888, 893, 895, 897, 902, 903, 904, 906, 907, 909, 910, 911, 912, 914, 916, 917, 918, 919, 921, 923, 925, 929, 930, 932
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			754 is a term because 754 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3.
		

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, 10):
        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])

A345862 Numbers that are the sum of ten fourth powers in exactly ten ways.

Original entry on oeis.org

6885, 7990, 8035, 8165, 8275, 8340, 8515, 8565, 8580, 9140, 9235, 9285, 9445, 9495, 9510, 9540, 9620, 9670, 9795, 9830, 9860, 9924, 9925, 9990, 10005, 10164, 10294, 10340, 10374, 10404, 10420, 10515, 10534, 10884, 10950, 10980, 11075, 11125, 11190, 11220
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345603 at term 4 because 8100 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 6^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 9^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 9^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Examples

			7990 is a term because 7990 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 1^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 9^4 = 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4.
		

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, 10):
        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])

A345642 Numbers that are the sum of ten fifth powers in ten or more ways.

Original entry on oeis.org

1431641, 1439416, 1464377, 1464408, 1505660, 1531398, 1531640, 1564165, 1697929, 1703935, 1782171, 1969222, 1969253, 1969464, 1976997, 1985183, 1986028, 2000966, 2001989, 2028270, 2042460, 2052415, 2058421, 2059202, 2060522, 2069221, 2076393, 2130272, 2182162
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1439416 is a term because 1439416 = 1^5 + 1^5 + 2^5 + 7^5 + 8^5 + 8^5 + 10^5 + 12^5 + 12^5 + 15^5 = 1^5 + 2^5 + 3^5 + 6^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 3^5 + 6^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 1^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 8^5 + 9^5 + 12^5 + 16^5 = 2^5 + 2^5 + 3^5 + 4^5 + 6^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 3^5 + 6^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 7^5 + 11^5 + 11^5 + 16^5 = 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**5 for x in range(1, 1000)]
    for pos in cwr(power_terms, 10):
        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 1-5 of 5 results.