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.

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])

A345818 Numbers that are the sum of six fourth powers in exactly six ways.

Original entry on oeis.org

37811, 38051, 43251, 43571, 44115, 44531, 45155, 45651, 45891, 47411, 47586, 49971, 52195, 53235, 54131, 56290, 57395, 57460, 57570, 59075, 59330, 59860, 60035, 62180, 62211, 63971, 66340, 67026, 67635, 67715, 67860, 67940, 68115, 68291, 68484, 69395, 69410
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345563 at term 1 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

			37811 is a term because 37811 = 1^4 + 2^4 + 2^4 + 7^4 + 11^4 + 12^4 = 2^4 + 2^4 + 4^4 + 7^4 + 9^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 9^4 + 13^4 = 3^4 + 4^4 + 8^4 + 8^4 + 11^4 + 11^4 = 4^4 + 6^4 + 7^4 + 9^4 + 9^4 + 12^4 = 5^4 + 5^4 + 9^4 + 10^4 + 10^4 + 10^4.

Links

Crossrefs

Cf. A344941, A345563, A345768, A345817, A345819, A345828, A346361.

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

A345175 Numbers that are the sum of five third powers in exactly six ways.

Original entry on oeis.org

2430, 2979, 3214, 3249, 3312, 3492, 3520, 3737, 3753, 3788, 3816, 3842, 3942, 3968, 4121, 4185, 4213, 4267, 4355, 4411, 4418, 4446, 4453, 4456, 4465, 4482, 4509, 4563, 4626, 4663, 4670, 4723, 4753, 4896, 4905, 4924, 4938, 4941, 4950, 4960, 4976, 4987, 4994
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Comments

Differs from A345174 at term 20 because 4392 = 1^3 + 1^3 + 10^3 + 10^3 + 11^3 = 1^3 + 2^3 + 2^3 + 9^3 + 14^3 = 1^3 + 8^3 + 9^3 + 10^3 + 10^3 = 2^3 + 2^3 + 3^3 + 5^3 + 15^3 = 2^3 + 3^3 + 5^3 + 8^3 + 14^3 = 2^3 + 8^3 + 8^3 + 8^3 + 12^3 = 3^3 + 6^3 + 7^3 + 8^3 + 13^3 = 5^3 + 5^3 + 5^3 + 9^3 + 13^3.

Examples

			2430 is a term because 2430 = 1^3 + 2^3 + 2^3 + 5^3 + 12^3  = 1^3 + 3^3 + 4^3 + 7^3 + 11^3  = 2^3 + 2^3 + 6^3 + 6^3 + 11^3  = 2^3 + 3^3 + 3^3 + 9^3 + 10^3  = 3^3 + 5^3 + 8^3 + 8^3 + 8^3  = 3^3 + 4^3 + 7^3 + 8^3 + 9^3.

Links

Crossrefs

Cf. A294740, A343988, A344941, A345149, A345174, A345181, 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, 5):
    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])

A345767 Numbers that are the sum of six cubes in exactly five ways.

Original entry on oeis.org

1045, 1169, 1241, 1260, 1384, 1432, 1440, 1495, 1530, 1539, 1549, 1556, 1558, 1584, 1594, 1602, 1612, 1617, 1640, 1654, 1657, 1675, 1703, 1712, 1715, 1719, 1729, 1736, 1745, 1747, 1754, 1771, 1780, 1792, 1801, 1803, 1806, 1810, 1818, 1825, 1827, 1834, 1843
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345514 at term 5 because 1377 = 1^3 + 1^3 + 2^3 + 7^3 + 8^3 + 8^3 = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 10^3 = 1^3 + 2^3 + 3^3 + 5^3 + 6^3 + 10^3 = 1^3 + 6^3 + 6^3 + 6^3 + 6^3 + 8^3 = 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 8^3 = 3^3 + 4^3 + 5^3 + 6^3 + 6^3 + 9^3.

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. A343988, A345514, A345766, A345768, A345777, A345817.

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])

A345769 Numbers that are the sum of six cubes in exactly seven ways.

Original entry on oeis.org

1710, 1766, 1773, 1988, 2051, 2160, 2196, 2249, 2251, 2259, 2314, 2322, 2349, 2375, 2417, 2424, 2480, 2492, 2513, 2520, 2531, 2539, 2548, 2564, 2565, 2574, 2611, 2613, 2639, 2656, 2702, 2707, 2762, 2770, 2773, 2792, 2798, 2808, 2818, 2825, 2826, 2833, 2844
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345516 at term 4 because 1981 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 12^3 = 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 12^3 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 10^3 = 1^3 + 1^3 + 6^3 + 6^3 + 6^3 + 11^3 = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 10^3 = 3^3 + 3^3 + 7^3 + 7^3 + 8^3 + 9^3 = 3^3 + 4^3 + 6^3 + 6^3 + 9^3 + 9^3 = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 10^3.

Examples

			1766 is a term because 1766 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 11^3 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 9^3 = 1^3 + 3^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 9^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3.

Links

Crossrefs

Cf. A345181, A345516, A345768, A345770, A345779, A345819.

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 == 7])
    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])
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.