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.

A345525 Numbers that are the sum of seven cubes in seven or more ways.

Original entry on oeis.org

1072, 1170, 1235, 1261, 1268, 1305, 1385, 1392, 1396, 1411, 1440, 1441, 1448, 1450, 1459, 1489, 1496, 1502, 1504, 1513, 1515, 1538, 1540, 1547, 1552, 1557, 1559, 1564, 1565, 1566, 1567, 1576, 1585, 1587, 1592, 1593, 1594, 1600, 1602, 1603, 1606, 1613, 1620
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345484, A345516, A345524, A345526, A345537, A345573, A345779.

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])

A345536 Numbers that are the sum of eight cubes in six or more 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, 902, 907, 908, 912, 921, 927, 928, 931, 938, 945, 946, 947, 951, 954, 956, 958, 963, 964, 965, 970, 972, 977, 982, 984, 989, 991, 992, 996, 998, 999, 1001
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345493, A345524, A345535, A345537, A345545, A345581, A345788.

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])

A345538 Numbers that are the sum of eight cubes in eight or more ways.

Original entry on oeis.org

970, 977, 984, 1054, 1073, 1075, 1080, 1090, 1099, 1106, 1110, 1125, 1129, 1136, 1148, 1160, 1166, 1171, 1178, 1181, 1185, 1186, 1188, 1192, 1197, 1204, 1206, 1211, 1217, 1218, 1223, 1225, 1230, 1232, 1234, 1236, 1237, 1242, 1243, 1249, 1262, 1263, 1269, 1273
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345495, A345526, A345537, A345539, A345547, A345583, A345790.

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])

A345546 Numbers that are the sum of nine cubes in seven or more ways.

Original entry on oeis.org

624, 629, 631, 650, 657, 687, 694, 707, 713, 720, 727, 744, 746, 753, 755, 763, 768, 770, 777, 779, 781, 784, 786, 789, 792, 796, 798, 803, 805, 807, 818, 820, 822, 824, 831, 833, 840, 842, 844, 847, 848, 849, 854, 859, 861, 866, 868, 870, 873, 875, 876, 877
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345504, A345537, A345545, A345547, A345555, A345591, A345799.

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])

A345582 Numbers that are the sum of eight fourth powers in seven or more ways.

Original entry on oeis.org

8003, 8243, 9043, 9218, 9283, 9523, 10372, 10803, 10868, 10948, 11043, 11412, 11557, 11587, 12083, 12692, 12932, 13188, 13268, 13333, 13508, 13972, 14147, 14212, 14387, 14788, 14883, 14933, 14948, 14963, 15013, 15028, 15093, 15173, 15268, 15317, 15332, 15397
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345537, A345573, A345581, A345583, A345591, A345615, A345839.

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])

A345789 Numbers that are the sum of eight cubes in exactly seven ways.

Original entry on oeis.org

902, 908, 921, 938, 958, 963, 982, 991, 996, 1003, 1008, 1010, 1017, 1019, 1028, 1029, 1033, 1047, 1055, 1058, 1061, 1062, 1070, 1087, 1091, 1094, 1096, 1097, 1104, 1108, 1111, 1113, 1115, 1116, 1118, 1120, 1122, 1123, 1127, 1134, 1141, 1143, 1145, 1152, 1153
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345537 at term 7 because 970 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 8^3 = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 9^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 + 7^3 = 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 8^3.
Likely finite.

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. A345537, A345779, A345788, A345790, A345799, A345839.

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])

A345494 Numbers that are the sum of eight squares in seven or more ways.

Original entry on oeis.org

56, 59, 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, 121
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			59 is a term because 59 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2.

Links

Crossrefs

Cf. A345484, A345493, A345495, A345504, A345537.

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

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

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