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.

A345515 Numbers that are the sum of six cubes in six or more ways.

Original entry on oeis.org

1377, 1488, 1586, 1595, 1647, 1673, 1677, 1710, 1738, 1764, 1766, 1773, 1799, 1829, 1836, 1837, 1862, 1881, 1890, 1911, 1953, 1955, 1981, 1988, 2007, 2011, 2014, 2018, 2025, 2044, 2051, 2070, 2079, 2097, 2105, 2107, 2108, 2142, 2153, 2160, 2168, 2170, 2177
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A344810, A345174, A345514, A345516, A345524, A345563, A345768.

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])

A345523 Numbers that are the sum of seven cubes in five or more ways.

Original entry on oeis.org

627, 768, 838, 845, 857, 864, 874, 881, 894, 900, 920, 937, 950, 955, 962, 969, 976, 981, 983, 990, 1002, 1009, 1011, 1016, 1027, 1046, 1053, 1054, 1060, 1061, 1063, 1072, 1079, 1089, 1096, 1098, 1102, 1105, 1107, 1109, 1115, 1117, 1121, 1124, 1128, 1133
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345482, A345514, A345522, A345524, A345535, A345571, A345777.

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])

A345525 Numbers that are the sum of seven cubes in seven or more ways.

Original entry on oeis.org

1072, 1170, 1235, 1261, 1268, 1305, 1385, 1392, 1396, 1411, 1440, 1441, 1448, 1450, 1459, 1489, 1496, 1502, 1504, 1513, 1515, 1538, 1540, 1547, 1552, 1557, 1559, 1564, 1565, 1566, 1567, 1576, 1585, 1587, 1592, 1593, 1594, 1600, 1602, 1603, 1606, 1613, 1620
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345484, A345516, A345524, A345526, A345537, A345573, A345779.

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])

A345536 Numbers that are the sum of eight cubes in six or more 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, 902, 907, 908, 912, 921, 927, 928, 931, 938, 945, 946, 947, 951, 954, 956, 958, 963, 964, 965, 970, 972, 977, 982, 984, 989, 991, 992, 996, 998, 999, 1001
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345493, A345524, A345535, A345537, A345545, A345581, A345788.

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])

A345572 Numbers that are the sum of seven fourth powers in six or more ways.

Original entry on oeis.org

10787, 15396, 15411, 15586, 15651, 16611, 16626, 16676, 16691, 16866, 17347, 17956, 17971, 18867, 19156, 19236, 19251, 19411, 19426, 19491, 19666, 20035, 20706, 20771, 21012, 21187, 21252, 21267, 21332, 21397, 21412, 21442, 21492, 21507, 21572, 21621, 21636
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345524, A345563, A345571, A345573, A345581, A345609, A345828.

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])

A345778 Numbers that are the sum of seven cubes in exactly six ways.

Original entry on oeis.org

955, 969, 1046, 1053, 1079, 1107, 1117, 1121, 1158, 1161, 1177, 1184, 1196, 1198, 1216, 1222, 1242, 1254, 1272, 1280, 1287, 1291, 1294, 1297, 1298, 1310, 1324, 1350, 1351, 1355, 1366, 1369, 1376, 1378, 1388, 1403, 1404, 1415, 1417, 1418, 1422, 1433, 1437
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345524 at term 5 because 1072 = 1^3 + 1^3 + 1^3 + 5^3 + 6^3 + 6^3 + 8^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 5^3 + 9^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 9^3 = 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 7^3 + 7^3 = 3^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3.
Likely finite.

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. A345524, A345768, A345777, A345779, A345788, A345828.

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])

A345483 Numbers that are the sum of seven squares in six or more ways.

Original entry on oeis.org

55, 58, 61, 63, 64, 66, 67, 69, 70, 71, 72, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			58 is a term because 58 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 = 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2.

Links

Crossrefs

Cf. A344810, A345482, A345484, A345493, A345524.

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, 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])

Formula

Conjectures from Chai Wah Wu, Apr 25 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 9.
G.f.: x*(-x^8 + x^7 - x^6 + x^5 - x^4 - x^3 - 52*x + 55)/(x - 1)^2. (End)
Showing 1-7 of 7 results.

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