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.

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])

A345839 Numbers that are the sum of eight fourth powers in exactly seven ways.

Original entry on oeis.org

8003, 8243, 9043, 9218, 9283, 9523, 10372, 10803, 10868, 10948, 11043, 11412, 11557, 11587, 12083, 12692, 12932, 13188, 13333, 13508, 13972, 14147, 14387, 14883, 14933, 14948, 14963, 15013, 15028, 15093, 15173, 15268, 15317, 15332, 15397, 15412, 15413, 15492
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345582 at term 19 because 13268 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 8^4 + 9^4 = 1^4 + 1^4 + 2^4 + 4^4 + 7^4 + 7^4 + 8^4 + 8^4 = 1^4 + 1^4 + 3^4 + 6^4 + 6^4 + 7^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 9^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 9^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 9^4 = 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 10^4.

Examples

			8243 is a term because 8243 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 8^4 = 2^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 3^4 + 4^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4.

Links

Crossrefs

Cf. A345582, A345789, A345829, A345838, A345840, A345849, A346332.

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

A345779 Numbers that are the sum of seven cubes in exactly seven ways.

Original entry on oeis.org

1072, 1170, 1235, 1261, 1268, 1305, 1392, 1396, 1411, 1440, 1441, 1448, 1450, 1459, 1489, 1502, 1504, 1513, 1538, 1540, 1547, 1559, 1564, 1565, 1566, 1567, 1576, 1592, 1593, 1594, 1600, 1602, 1606, 1620, 1621, 1625, 1626, 1628, 1629, 1639, 1658, 1664, 1667
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345525 at term 7 because 1385 = 1^3 + 1^3 + 2^3 + 2^3 + 7^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 10^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 10^3 = 1^3 + 2^3 + 6^3 + 6^3 + 6^3 + 6^3 + 8^3 = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 8^3 = 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 6^3 + 9^3 = 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 8^3 + 8^3.
Likely finite.

Examples

			1170 is a term because 1170 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3 = 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 7^3.

Links

Crossrefs

Cf. A345525, A345769, A345778, A345780, A345789, A345829.

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 == 7])
    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])

A345790 Numbers that are the sum of eight cubes in exactly eight ways.

Original entry on oeis.org

970, 977, 1054, 1073, 1075, 1090, 1099, 1106, 1110, 1125, 1129, 1148, 1160, 1166, 1178, 1181, 1186, 1188, 1206, 1211, 1217, 1218, 1225, 1230, 1232, 1234, 1236, 1237, 1242, 1249, 1263, 1276, 1281, 1286, 1292, 1298, 1305, 1312, 1314, 1321, 1323, 1324, 1334
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345538 at term 3 because 984 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 9^3 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 = 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 6^3 + 7^3 + 7^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3.
Likely finite.

Examples

			977 is a term because 977 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 8^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A345538, A345780, A345789, A345791, A345800, A345840.

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 == 8])
    for x in range(len(rets)):
        print(rets[x])

A345799 Numbers that are the sum of nine cubes in exactly seven ways.

Original entry on oeis.org

624, 629, 631, 650, 657, 687, 694, 707, 713, 720, 727, 746, 753, 755, 763, 768, 777, 779, 781, 784, 786, 789, 792, 796, 798, 803, 807, 820, 822, 824, 831, 833, 848, 849, 854, 870, 873, 875, 876, 879, 884, 885, 889, 890, 892, 898, 899, 901, 902, 904, 905, 906
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345546 at term 12 because 744 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 7^3 = 1^3 + 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 + 4^3 + 4^3 + 4^3 + 8^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3.
Likely finite.

Examples

			629 is a term because 629 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 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 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3.

Links

Crossrefs

Cf. A345546, A345789, A345798, A345800, A345809, A345849.

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 == 7])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-6 of 6 results.

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

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