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.

A345539 Numbers that are the sum of eight cubes in nine or more ways.

Original entry on oeis.org

984, 1080, 1136, 1171, 1185, 1192, 1197, 1204, 1223, 1243, 1262, 1269, 1273, 1280, 1288, 1295, 1299, 1306, 1318, 1325, 1332, 1333, 1337, 1344, 1356, 1360, 1369, 1370, 1374, 1377, 1379, 1386, 1393, 1397, 1400, 1404, 1406, 1412, 1415, 1416, 1419, 1422, 1423
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1080 is a term because 1080 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 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, 8):
        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])

A345547 Numbers that are the sum of nine cubes in eight or more ways.

Original entry on oeis.org

744, 770, 805, 818, 840, 842, 844, 847, 859, 861, 866, 868, 877, 880, 883, 887, 894, 896, 903, 908, 909, 910, 911, 913, 915, 916, 920, 922, 929, 935, 939, 940, 945, 946, 948, 950, 952, 954, 955, 957, 959, 961, 964, 965, 966, 971, 972, 973, 976, 978, 983, 985
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			770 is a term because 770 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 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, 9):
        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])

A345593 Numbers that are the sum of nine fourth powers in nine or more ways.

Original entry on oeis.org

8259, 9299, 9539, 10709, 10819, 10884, 10949, 10964, 11124, 11444, 11573, 11668, 11684, 11924, 12099, 12164, 12339, 12404, 12549, 12708, 12773, 12853, 12918, 12948, 13013, 13139, 13204, 13269, 13284, 13349, 13379, 13444, 13509, 13524, 13589, 13764, 13829
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			9299 is a term because 9299 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 7^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 9^4 = 3^4 + 3^4 + 4^4 + 6^4 + 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, 9):
        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])

A345549 Numbers that are the sum of nine cubes in ten or more ways.

Original entry on oeis.org

966, 971, 978, 985, 992, 1004, 1011, 1018, 1022, 1048, 1055, 1056, 1062, 1063, 1074, 1076, 1078, 1081, 1083, 1085, 1088, 1092, 1093, 1095, 1097, 1098, 1100, 1102, 1104, 1107, 1109, 1111, 1112, 1114, 1117, 1118, 1119, 1121, 1123, 1124, 1126, 1128, 1130, 1133
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			971 is a term because 971 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 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, 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])

A345801 Numbers that are the sum of nine cubes in exactly nine ways.

Original entry on oeis.org

859, 861, 896, 903, 922, 929, 935, 939, 959, 973, 997, 999, 1009, 1016, 1020, 1023, 1027, 1029, 1030, 1034, 1035, 1036, 1037, 1041, 1046, 1059, 1060, 1064, 1065, 1066, 1067, 1071, 1072, 1079, 1086, 1091, 1105, 1116, 1131, 1138, 1146, 1152, 1155, 1157, 1158
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345548 at term 10 because 966 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 6^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 8^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 + 7^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 9^3 = 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3 = 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3.
Likely finite.

Examples

			861 is a term because 861 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 8^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3.
		

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, 9):
        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])

A345557 Numbers that are the sum of ten cubes in nine or more ways.

Original entry on oeis.org

632, 651, 658, 688, 695, 714, 721, 736, 740, 745, 752, 754, 756, 771, 773, 778, 780, 782, 790, 792, 795, 797, 799, 801, 806, 808, 812, 813, 815, 816, 818, 819, 821, 823, 825, 832, 834, 841, 843, 845, 847, 848, 849, 850, 851, 852, 853, 855, 856, 857, 858, 860
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			651 is a term because 651 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 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 >= 9])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.