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.

Links

Crossrefs

Cf. A345496, A345527, A345538, A345540, A345548, A345584, A345791.

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])

A345841 Numbers that are the sum of eight fourth powers in exactly nine ways.

Original entry on oeis.org

15427, 16692, 17348, 17493, 18052, 18227, 19267, 19412, 19572, 19748, 20852, 21443, 21493, 21637, 21652, 21653, 21827, 21877, 21972, 22037, 22212, 22388, 22501, 22548, 22868, 22932, 23107, 23412, 23413, 23428, 23828, 23893, 23972, 24037, 24131, 24212, 24517
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345584 at term 5 because 17972 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 + 10^4 = 1^4 + 1^4 + 5^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 5^4 + 7^4 + 11^4 = 1^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 6^4 + 11^4 = 1^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 + 10^4 = 1^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 + 10^4 = 1^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 8^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 7^4 + 9^4 + 9^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 9^4 + 9^4.

Examples

			16692 is a term because 16692 = 1^4 + 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 + 10^4 = 1^4 + 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 11^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 10^4 = 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 1^4 + 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 5^4 + 7^4 + 9^4 + 9^4 = 2^4 + 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 9^4 + 9^4.

Links

Crossrefs

Cf. A345584, A345791, A345831, A345840, A345842, A345851, A346334.

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 == 9])
    for x in range(len(rets)):
        print(rets[x])

A345781 Numbers that are the sum of seven cubes in exactly nine ways.

Original entry on oeis.org

1496, 1648, 1720, 1737, 1772, 1781, 1802, 1835, 1844, 1882, 1891, 1898, 1900, 1907, 1912, 1919, 1945, 1952, 1954, 1961, 1996, 2000, 2012, 2026, 2071, 2080, 2098, 2107, 2110, 2115, 2116, 2132, 2134, 2136, 2139, 2150, 2152, 2168, 2178, 2185, 2187, 2195, 2205
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345527 at term 3 because 1704 = 1^3 + 1^3 + 1^3 + 3^3 + 6^3 + 9^3 + 9^3 = 1^3 + 1^3 + 1^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 1^3 + 2^3 + 2^3 + 7^3 + 7^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 6^3 + 11^3 = 1^3 + 2^3 + 4^3 + 6^3 + 7^3 + 7^3 + 9^3 = 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 6^3 + 11^3 = 2^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 7^3 + 10^3 = 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 + 9^3 = 3^3 + 6^3 + 6^3 + 6^3 + 7^3 + 7^3 + 7^3.
Likely finite.

Examples

			1648 is a term because 1648 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 5^3 + 6^3 + 6^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 + 8^3 = 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 7^3 + 7^3.

Links

Crossrefs

Cf. A345527, A345771, A345780, A345782, A345791, A345831.

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 == 9])
    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])

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.

Links

Crossrefs

Cf. A345548, A345791, A345800, A345802, A345811, A345851.

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])

A345792 Numbers that are the sum of eight cubes in exactly ten ways.

Original entry on oeis.org

1185, 1243, 1288, 1295, 1299, 1386, 1397, 1400, 1412, 1423, 1448, 1449, 1451, 1458, 1460, 1464, 1467, 1475, 1477, 1501, 1503, 1505, 1512, 1513, 1516, 1539, 1540, 1541, 1553, 1558, 1559, 1568, 1577, 1578, 1586, 1588, 1591, 1592, 1594, 1595, 1596, 1600, 1608
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345540 at term 3 because 1262 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 6^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 7^3 + 7^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 6^3 + 9^3 = 1^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 6^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 10^3 = 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 9^3.
Likely finite.

Examples

			1243 is a term because 1243 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 8^3 = 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 7^3.

Links

Crossrefs

Cf. A345540, A345782, A345791, A345802, A345842.

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 == 10])
    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.

Content is available under The OEIS End-User License Agreement.