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 1-7 of 7 results.

A345524 Numbers that are the sum of seven cubes in six or more ways.

Original entry on oeis.org

955, 969, 1046, 1053, 1072, 1079, 1107, 1117, 1121, 1158, 1161, 1170, 1177, 1184, 1196, 1198, 1216, 1222, 1235, 1242, 1254, 1261, 1268, 1272, 1280, 1287, 1291, 1294, 1297, 1298, 1305, 1310, 1324, 1350, 1351, 1355, 1366, 1369, 1376, 1378, 1385, 1388, 1392
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			969 is a term because 969 = 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A345483, A345515, A345523, A345525, A345536, A345572, A345778.

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])

A345535 Numbers that are the sum of eight cubes in five or more ways.

Original entry on oeis.org

471, 497, 504, 597, 623, 628, 630, 635, 642, 649, 654, 661, 667, 680, 686, 691, 693, 712, 717, 719, 723, 728, 736, 738, 741, 743, 752, 754, 755, 762, 769, 774, 776, 778, 780, 781, 783, 784, 785, 788, 791, 793, 795, 797, 800, 802, 804, 810, 813, 814, 815, 817
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			497 is a term because 497 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.

Links

Crossrefs

Cf. A345492, A345523, A345534, A345536, A345544, A345580, A345787.

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])

A345537 Numbers that are the sum of eight cubes in seven or more ways.

Original entry on oeis.org

902, 908, 921, 938, 958, 963, 970, 977, 982, 984, 991, 996, 1003, 1008, 1010, 1017, 1019, 1028, 1029, 1033, 1047, 1054, 1055, 1058, 1061, 1062, 1070, 1073, 1075, 1080, 1087, 1090, 1091, 1094, 1096, 1097, 1099, 1104, 1106, 1108, 1110, 1111, 1113, 1115, 1116
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			908 is a term because 908 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3.

Links

Crossrefs

Cf. A345494, A345525, A345536, A345538, A345546, A345582, A345789.

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 >= 7])
    for x in range(len(rets)):
        print(rets[x])

A345545 Numbers that are the sum of nine cubes in six or more ways.

Original entry on oeis.org

472, 498, 505, 507, 524, 596, 598, 605, 624, 629, 631, 636, 643, 650, 655, 657, 661, 662, 669, 672, 676, 681, 687, 688, 690, 692, 694, 696, 706, 707, 713, 718, 720, 722, 725, 727, 728, 729, 731, 732, 737, 739, 742, 744, 746, 748, 749, 750, 751, 753, 755, 756
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			498 is a term because 498 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3.

Links

Crossrefs

Cf. A345503, A345536, A345544, A345546, A345554, A345590, A345798.

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])

A345581 Numbers that are the sum of eight fourth powers in six or more ways.

Original entry on oeis.org

6723, 6788, 6853, 6898, 6963, 7028, 7938, 8003, 8068, 8178, 8243, 8308, 8483, 8963, 9043, 9173, 9218, 9283, 9348, 9413, 9493, 9523, 9668, 9763, 9828, 10003, 10132, 10258, 10277, 10307, 10372, 10628, 10708, 10738, 10788, 10803, 10868, 10933, 10948, 10978
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6788 is a term because 6788 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.

Links

Crossrefs

Cf. A345536, A345572, A345580, A345582, A345590, A345614, A345838.

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, 8):
    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])

A345788 Numbers that are the sum of eight cubes in exactly six ways.

Original entry on oeis.org

628, 719, 769, 776, 778, 795, 832, 839, 846, 858, 860, 865, 872, 875, 876, 882, 886, 891, 893, 895, 901, 907, 912, 927, 928, 931, 945, 946, 947, 951, 954, 956, 964, 965, 972, 989, 992, 998, 999, 1001, 1012, 1014, 1015, 1021, 1034, 1035, 1036, 1038, 1040, 1045
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345536 at term 22 because 902 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 9^3 = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 5^3 + 6^3 + 7^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 7^3 = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3 = 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3 = 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3.
Likely finite.

Examples

			719 is a term because 719 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3.

Links

Crossrefs

Cf. A345536, A345778, A345787, A345789, A345798, A345838.

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 == 6])
    for x in range(len(rets)):
        print(rets[x])

A345493 Numbers that are the sum of eight squares in six or more ways.

Original entry on oeis.org

56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			58 is a term because 58 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 5^2.

Links

Crossrefs

Cf. A345483, A345492, A345494, A345503, A345536.

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, 8):
    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])

Formula

Conjectures from Chai Wah Wu, Apr 25 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 11.
G.f.: x*(-x^10 + x^9 - x^8 + x^7 - x^6 + x^5 - x^4 + x^3 - x^2 - 54*x + 56)/(x - 1)^2. (End)
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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