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.

A345524 Numbers that are the sum of seven cubes in six or more ways.

Original entry on oeis.org

955, 969, 1046, 1053, 1072, 1079, 1107, 1117, 1121, 1158, 1161, 1170, 1177, 1184, 1196, 1198, 1216, 1222, 1235, 1242, 1254, 1261, 1268, 1272, 1280, 1287, 1291, 1294, 1297, 1298, 1305, 1310, 1324, 1350, 1351, 1355, 1366, 1369, 1376, 1378, 1385, 1388, 1392
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			969 is a term because 969 = 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A345483, A345515, A345523, A345525, A345536, A345572, A345778.

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])

A345828 Numbers that are the sum of seven fourth powers in exactly six ways.

Original entry on oeis.org

10787, 15396, 15411, 15586, 15651, 16611, 16626, 16676, 16866, 17956, 18867, 19156, 19236, 19251, 19411, 19426, 19666, 20035, 20771, 21012, 21187, 21397, 21412, 21442, 21492, 21572, 21621, 21811, 21891, 22116, 22132, 22292, 22307, 22372, 22595, 22660, 22962
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345572 at term 9 because 16691 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4.

Examples

			15396 is a term because 15396 = 1^4 + 1^4 + 1^4 + 1^4 + 6^4 + 8^4 + 10^4 = 1^4 + 1^4 + 2^4 + 5^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 11^4 = 1^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 10^4 = 1^4 + 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 3^4 + 4^4 + 5^4 + 6^4 + 9^4 + 9^4.

Links

Crossrefs

Cf. A345572, A345778, A345818, A345827, A345829, A345838, A346283.

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, 7):
    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])

A345768 Numbers that are the sum of six cubes in exactly six ways.

Original entry on oeis.org

1377, 1488, 1586, 1595, 1647, 1673, 1677, 1738, 1764, 1799, 1829, 1836, 1837, 1862, 1881, 1890, 1911, 1953, 1955, 2007, 2011, 2014, 2018, 2025, 2044, 2070, 2079, 2097, 2107, 2108, 2142, 2153, 2170, 2177, 2203, 2214, 2216, 2222, 2223, 2226, 2229, 2252, 2258
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345515 at term 8 because 1710 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 6^3 + 11^3 = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.

Examples

			1488 is a term because 1488 = 1^3 + 1^3 + 1^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A345175, A345515, A345767, A345769, A345778, A345818.

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, 6):
    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])

A345777 Numbers that are the sum of seven cubes in exactly five ways.

Original entry on oeis.org

627, 768, 838, 845, 857, 864, 874, 881, 894, 900, 920, 937, 950, 962, 976, 981, 983, 990, 1002, 1009, 1011, 1016, 1027, 1054, 1060, 1061, 1063, 1089, 1096, 1098, 1102, 1105, 1109, 1115, 1124, 1128, 1133, 1135, 1137, 1140, 1144, 1151, 1153, 1154, 1159, 1163
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345523 at term 14 because 955 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3 + 6^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 9^3 = 1^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 + 7^3 = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 8^3 = 2^3 + 2^3 + 4^3 + 4^3 + 5^3 + 7^3 + 7^3 = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 8^3.
Likely finite.

Examples

			768 is a term because 768 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3.

Links

Crossrefs

Cf. A345523, A345767, A345776, A345778, A345787, A345827.

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 == 5])
    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])
Showing 1-6 of 6 results.

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