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.

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])

A345826 Numbers that are the sum of seven fourth powers in exactly four ways.

Original entry on oeis.org

2932, 4147, 4212, 4387, 5427, 5602, 5667, 6627, 6692, 6817, 6822, 6837, 6852, 6867, 7012, 7122, 7251, 7316, 7491, 7747, 7857, 8052, 8097, 8162, 8402, 8467, 8532, 8707, 8787, 9027, 9092, 9157, 9172, 9202, 9237, 9252, 9332, 9412, 9442, 9492, 9572, 9652, 9682
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345570 at term 9 because 6642 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Examples

			4147 is a term because 4147 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4.

Links

Crossrefs

Cf. A345570, A345776, A345816, A345825, A345827, A345836, A346281.

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 == 4])
    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])

A345775 Numbers that are the sum of seven cubes in exactly three ways.

Original entry on oeis.org

222, 229, 248, 255, 262, 281, 283, 285, 318, 346, 370, 374, 377, 379, 381, 396, 400, 407, 412, 419, 426, 433, 437, 438, 444, 451, 463, 472, 475, 477, 489, 494, 501, 505, 507, 510, 522, 529, 533, 536, 559, 564, 566, 568, 570, 577, 578, 584, 585, 592, 594, 596
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345521 at term 28 because 470 = 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 = 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3.
Likely finite.

Examples

			229 is a term because 229 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Links

Crossrefs

Cf. A048931, A345521, A345774, A345776, A345785, A345825.

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 == 3])
    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])

A345786 Numbers that are the sum of eight cubes in exactly four ways.

Original entry on oeis.org

256, 347, 382, 401, 408, 427, 434, 438, 445, 464, 478, 480, 490, 499, 502, 506, 511, 516, 523, 530, 532, 534, 537, 560, 565, 567, 569, 571, 578, 586, 593, 595, 600, 602, 604, 605, 611, 612, 616, 619, 621, 624, 626, 643, 645, 656, 660, 663, 664, 668, 675, 679
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345534 at term 11 because 471 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3.
Likely finite.

Examples

			347 is a term because 347 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3.

Links

Crossrefs

Cf. A345534, A345776, A345785, A345787, A345796, A345836.

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 == 4])
    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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