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.

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.

Links

Crossrefs

Cf. A345540, A345548, A345558, A345594, A345802, A346803.

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])

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.

Links

Crossrefs

Cf. A345548, A345556, A345558, A345602, A345811.

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])

A345603 Numbers that are the sum of ten fourth powers in ten or more ways.

Original entry on oeis.org

6885, 7990, 8035, 8100, 8165, 8275, 8340, 8515, 8565, 8580, 9140, 9205, 9235, 9285, 9300, 9315, 9380, 9445, 9495, 9510, 9540, 9555, 9620, 9670, 9685, 9750, 9795, 9830, 9860, 9924, 9925, 9990, 10005, 10164, 10294, 10340, 10374, 10404, 10420, 10515, 10534
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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.

Links

Crossrefs

Cf. A345558, A345594, A345602, A345642, A345862.

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])

A345812 Numbers that are the sum of ten cubes in exactly ten ways.

Original entry on oeis.org

721, 754, 756, 782, 792, 797, 806, 808, 819, 834, 847, 848, 850, 860, 871, 874, 876, 877, 878, 881, 884, 886, 893, 902, 903, 907, 909, 910, 916, 917, 918, 921, 929, 932, 933, 936, 937, 938, 941, 942, 944, 945, 955, 965, 966, 968, 973, 991, 994, 999, 1001
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345558 at term 4 because 771 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 9^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 7^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3.
Likely finite.

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.

Links

Crossrefs

Cf. A345558, A345802, A345811, A345862.

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])

A346808 Numbers that are the sum of ten squares in ten or more ways.

Original entry on oeis.org

61, 64, 66, 67, 69, 70, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			64 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2
so 64 is a term.

Links

Crossrefs

Cf. A345558, A346803. Subsequence of A346807.

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, 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.

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