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.

A345526 Numbers that are the sum of seven cubes in eight or more ways.

Original entry on oeis.org

1385, 1496, 1515, 1552, 1557, 1585, 1587, 1603, 1613, 1622, 1648, 1655, 1665, 1674, 1681, 1704, 1711, 1718, 1719, 1720, 1737, 1739, 1741, 1746, 1753, 1755, 1765, 1767, 1772, 1774, 1781, 1782, 1793, 1800, 1802, 1805, 1809, 1811, 1818, 1819, 1826, 1828, 1830
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1496 is a term because 1496 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 2^3 + 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 + 7^3.

Links

Crossrefs

Cf. A345485, A345517, A345525, A345527, A345538, A345574, A345780.

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 >= 8])
    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])

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])

A345547 Numbers that are the sum of nine cubes in eight or more ways.

Original entry on oeis.org

744, 770, 805, 818, 840, 842, 844, 847, 859, 861, 866, 868, 877, 880, 883, 887, 894, 896, 903, 908, 909, 910, 911, 913, 915, 916, 920, 922, 929, 935, 939, 940, 945, 946, 948, 950, 952, 954, 955, 957, 959, 961, 964, 965, 966, 971, 972, 973, 976, 978, 983, 985
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			770 is a term because 770 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.

Links

Crossrefs

Cf. A345505, A345538, A345546, A345548, A345556, A345592, A345800.

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

A345583 Numbers that are the sum of eight fourth powers in eight or more ways.

Original entry on oeis.org

13268, 14212, 14788, 15427, 15667, 16612, 16627, 16692, 16707, 16772, 16822, 16852, 16882, 16947, 17348, 17363, 17428, 17493, 17877, 17972, 17987, 18052, 18117, 18227, 18948, 19157, 19237, 19252, 19267, 19412, 19492, 19507, 19572, 19682, 19747, 19748, 19828
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			14212 is a term because 14212 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 8^4 + 10^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 10^4 = 1^4 + 1^4 + 1^4 + 5^4 + 6^4 + 8^4 + 8^4 + 8^4 = 1^4 + 2^4 + 4^4 + 4^4 + 5^4 + 7^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 8^4 + 9^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 10^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 10^4 = 3^4 + 4^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 8^4.

Links

Crossrefs

Cf. A345538, A345574, A345582, A345584, A345592, A345616, A345840.

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 >= 8])
    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])

A345495 Numbers that are the sum of eight squares in eight or more ways.

Original entry on oeis.org

56, 59, 62, 64, 65, 67, 68, 70, 71, 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, 122
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. A345485, A345494, A345496, A345505, A345538.

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-7 of 7 results.

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