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.

A345540 Numbers that are the sum of eight cubes in ten or more ways.

Original entry on oeis.org

1185, 1243, 1262, 1288, 1295, 1299, 1386, 1393, 1397, 1400, 1412, 1419, 1423, 1448, 1449, 1451, 1458, 1460, 1464, 1467, 1475, 1477, 1497, 1501, 1503, 1504, 1505, 1512, 1513, 1514, 1516, 1521, 1523, 1539, 1540, 1541, 1542, 1553, 1558, 1559, 1560, 1565, 1566
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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.
		

Crossrefs

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])

A345791 Numbers that are the sum of eight cubes in exactly nine ways.

Original entry on oeis.org

984, 1080, 1136, 1171, 1192, 1197, 1204, 1223, 1269, 1273, 1280, 1306, 1318, 1325, 1332, 1333, 1337, 1344, 1356, 1360, 1369, 1370, 1374, 1377, 1379, 1404, 1406, 1415, 1416, 1422, 1425, 1430, 1432, 1438, 1442, 1444, 1445, 1456, 1476, 1481, 1486, 1488, 1494
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345539 at term 5 because 1185 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 6^3 + 8^3 = 1^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 + 9^3 = 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 9^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 8^3 + 8^3 = 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 + 7^3 + 7^3.
Likely finite.

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.
		

Crossrefs

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])

A345842 Numbers that are the sum of eight fourth powers in exactly ten ways.

Original entry on oeis.org

17972, 17987, 19492, 19507, 19747, 20116, 21283, 21333, 21413, 21508, 21588, 22067, 22563, 23237, 23252, 23587, 23588, 23603, 23653, 24277, 24452, 24802, 24948, 25603, 26228, 27347, 27683, 27813, 27893, 27973, 28532, 28852, 28853, 28933, 29108, 29173, 29491
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345585 at term 7 because 20787 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 + 10^4 = 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 8^4 + 10^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4 + 11^4 = 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

Examples

			17987 is a term because 17987 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 5^4 + 7^4 + 11^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 + 10^4 = 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 + 10^4 = 2^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 + 10^4 = 3^4 + 5^4 + 6^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4.
		

Crossrefs

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 == 10])
        for x in range(len(rets)):
            print(rets[x])

A345782 Numbers that are the sum of seven cubes in exactly ten ways.

Original entry on oeis.org

1704, 1711, 1800, 1837, 1863, 1926, 1938, 1963, 2008, 2019, 2045, 2053, 2059, 2078, 2113, 2143, 2161, 2171, 2176, 2217, 2223, 2250, 2260, 2266, 2276, 2286, 2295, 2304, 2313, 2315, 2331, 2350, 2354, 2357, 2374, 2404, 2412, 2413, 2442, 2444, 2446, 2447, 2511
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345506 at term 3 because 1774 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 12^3 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3 + 6^3 + 11^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 9^3 + 10^3 = 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 11^3 = 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 + 9^3 = 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.
Likely finite.

Examples

			1711 is a term because 1711 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 9^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 4^3 + 4^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.
		

Crossrefs

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 == 10])
        for x in range(len(rets)):
            print(rets[x])

A345802 Numbers that are the sum of nine cubes in exactly ten ways.

Original entry on oeis.org

966, 971, 978, 1004, 1018, 1022, 1055, 1056, 1062, 1063, 1074, 1076, 1078, 1085, 1088, 1092, 1093, 1095, 1098, 1100, 1104, 1111, 1112, 1114, 1117, 1119, 1124, 1130, 1134, 1135, 1139, 1140, 1142, 1147, 1149, 1153, 1160, 1167, 1168, 1170, 1180, 1181, 1182, 1183
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345549 at term 4 because 985 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 9^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 6^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 8^3 = 1^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 9^3 = 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3.
Likely finite.

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.
		

Crossrefs

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])
Showing 1-5 of 5 results.