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.

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])

A345827 Numbers that are the sum of seven fourth powers in exactly five ways.

Original entry on oeis.org

6642, 6707, 6772, 6882, 6947, 7922, 7987, 8227, 8962, 9267, 9507, 9747, 10116, 10291, 10722, 10867, 10932, 10962, 11331, 11411, 11571, 12676, 12851, 12916, 13187, 13252, 13891, 13956, 14131, 14211, 14707, 14772, 14802, 14917, 14932, 14947, 15012, 15092, 15316
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345571 at term 16 because 10787 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 = 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 8^4.

Examples

			6707 is a term because 6707 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.

Links

Crossrefs

Cf. A345571, A345777, A345817, A345826, A345828, A345837, A346282.

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 == 5])
    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])

A345776 Numbers that are the sum of seven cubes in exactly four ways.

Original entry on oeis.org

470, 496, 503, 603, 634, 653, 659, 685, 690, 692, 711, 712, 747, 751, 754, 761, 766, 773, 775, 777, 780, 783, 787, 792, 794, 812, 813, 829, 831, 836, 842, 843, 859, 867, 871, 875, 883, 885, 890, 892, 899, 901, 904, 906, 907, 911, 913, 918, 919, 927, 930, 936
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345522 at term 5 because 627 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3.
Likely finite.

Examples

			496 is a term because 496 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3.

Links

Crossrefs

Cf. A345522, A345766, A345775, A345777, A345786, A345826.

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 == 4])
    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])

A345787 Numbers that are the sum of eight cubes in exactly five ways.

Original entry on oeis.org

471, 497, 504, 597, 623, 630, 635, 642, 649, 654, 661, 667, 680, 686, 691, 693, 712, 717, 723, 728, 736, 738, 741, 743, 752, 754, 755, 762, 774, 780, 781, 783, 784, 785, 788, 791, 793, 797, 800, 802, 804, 810, 813, 814, 815, 817, 819, 820, 821, 830, 834, 837
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345535 at term 6 because 628 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 + 7^3.
Likely finite.

Examples

			497 is a term because 497 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.

Links

Crossrefs

Cf. A345535, A345777, A345786, A345788, A345797, A345837.

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 == 5])
    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.

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