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.

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])

A345522 Numbers that are the sum of seven cubes in four or more ways.

Original entry on oeis.org

470, 496, 503, 603, 627, 634, 653, 659, 685, 690, 692, 711, 712, 747, 751, 754, 761, 766, 768, 773, 775, 777, 780, 783, 787, 792, 794, 812, 813, 829, 831, 836, 838, 842, 843, 845, 857, 859, 864, 867, 871, 874, 875, 881, 883, 885, 890, 892, 894, 899, 900, 901
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345481, A345513, A345521, A345523, A345534, A345570, A345776.

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])

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])

A345535 Numbers that are the sum of eight cubes in five or more ways.

Original entry on oeis.org

471, 497, 504, 597, 623, 628, 630, 635, 642, 649, 654, 661, 667, 680, 686, 691, 693, 712, 717, 719, 723, 728, 736, 738, 741, 743, 752, 754, 755, 762, 769, 774, 776, 778, 780, 781, 783, 784, 785, 788, 791, 793, 795, 797, 800, 802, 804, 810, 813, 814, 815, 817
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345492, A345523, A345534, A345536, A345544, A345580, A345787.

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])

A345571 Numbers that are the sum of seven fourth powers in five or more ways.

Original entry on oeis.org

6642, 6707, 6772, 6882, 6947, 7922, 7987, 8227, 8962, 9267, 9507, 9747, 10116, 10291, 10722, 10787, 10867, 10932, 10962, 11331, 11411, 11571, 12676, 12851, 12916, 13187, 13252, 13891, 13956, 14131, 14211, 14707, 14772, 14802, 14917, 14932, 14947, 15012, 15092
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345523, A345562, A345570, A345572, A345580, A345608, A345827.

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])

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])

A345482 Numbers that are the sum of seven squares in five or more ways.

Original entry on oeis.org

45, 54, 55, 57, 58, 60, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			54 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 5^2
so 54 is a term.

Links

Crossrefs

Cf. A344809, A345481, A345483, A345492, A345523.

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 >= 5])
    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 > 13.
G.f.: x*(-x^12 + x^11 - x^10 + x^9 - x^8 + x^7 - x^6 + x^5 - x^4 + x^3 - 8*x^2 - 36*x + 45)/(x - 1)^2. (End)
Showing 1-7 of 7 results.

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