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.

A345514 Numbers that are the sum of six cubes in five or more ways.

Original entry on oeis.org

1045, 1169, 1241, 1260, 1377, 1384, 1432, 1440, 1488, 1495, 1530, 1539, 1549, 1556, 1558, 1584, 1586, 1594, 1595, 1602, 1612, 1617, 1640, 1647, 1654, 1657, 1673, 1675, 1677, 1703, 1710, 1712, 1715, 1719, 1729, 1736, 1738, 1745, 1747, 1754, 1764, 1766, 1771
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1169 is a term because 1169 = 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3.

Links

Crossrefs

Cf. A343989, A344809, A345513, A345515, A345523, A345562, A345767.

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])

A345817 Numbers that are the sum of six fourth powers in exactly five ways.

Original entry on oeis.org

15395, 16610, 18866, 19235, 19410, 20996, 21011, 21316, 21331, 21491, 21620, 23811, 25091, 29700, 29715, 29906, 29955, 30356, 30995, 31235, 31266, 31331, 31506, 32035, 33651, 33795, 33891, 35171, 35411, 35636, 35796, 35971, 37971, 38595, 38675, 39266, 39890
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345562 at term 8 because 21251 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4.

Examples

			16610 is a term because 16610 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 5^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 10^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4.

Links

Crossrefs

Cf. A344359, A345562, A345767, A345816, A345818, A345827, A346360.

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, 6):
    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])

A343988 Numbers that are the sum of five positive cubes in exactly five ways.

Original entry on oeis.org

1765, 1980, 2043, 2104, 2195, 2250, 2449, 2486, 2491, 2493, 2547, 2584, 2592, 2738, 2745, 2764, 2817, 2888, 2915, 2953, 2969, 3095, 3096, 3133, 3142, 3186, 3188, 3240, 3275, 3277, 3310, 3366, 3403, 3422, 3459, 3464, 3466, 3483, 3529, 3583, 3608, 3627, 3653, 3664, 3671, 3690, 3697, 3707, 3725, 3744, 3746, 3781
Offset: 1

Views

Author

David Consiglio, Jr., May 06 2021

Keywords

Comments

Differs from A343989 at term 7 because 2430 = 1^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 4^3 + 5^3 + 8^3 + 12^3 = 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 4^3 + 5^3 + 8^3 + 9^3 + 10^3.

Examples

			2043 is a term because 2043 = 1^3 + 4^3 + 5^3 + 5^3 + 12^3 = 2^3 + 2^3 + 3^3 + 10^3 + 10^3 = 2^3 + 3^3 + 4^3 + 6^3 + 12^3 = 4^3 + 5^3 + 5^3 + 9^3 + 10^3 = 4^3 + 6^3 + 6^3 + 6^3 + 11^3.

Links

Crossrefs

Cf. A294739, A343986, A343989, A344035, A344359, A345175, A345767.

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(1,50)]
for pos in cwr(power_terms,5):
    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])

A345766 Numbers that are the sum of six cubes in exactly four ways.

Original entry on oeis.org

626, 830, 837, 856, 873, 891, 947, 954, 982, 1008, 1026, 1052, 1053, 1071, 1094, 1097, 1106, 1109, 1134, 1143, 1150, 1153, 1172, 1195, 1208, 1227, 1234, 1253, 1267, 1278, 1279, 1283, 1286, 1290, 1297, 1316, 1323, 1324, 1358, 1361, 1368, 1369, 1376, 1395, 1403
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345513 at term 12 because 1045 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 10^3 = 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 9^3 = 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 9^3 = 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 = 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3.

Examples

			830 is a term because 830 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 8^3 = 1^3 + 3^3 + 3^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A048931, A344035, A345513, A345767, A345776, A345816.

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 == 4])
    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])
Showing 1-6 of 6 results.

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