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.

A345534 Numbers that are the sum of eight cubes in four or more ways.

Original entry on oeis.org

256, 347, 382, 401, 408, 427, 434, 438, 445, 464, 471, 478, 480, 490, 497, 499, 502, 504, 506, 511, 516, 523, 530, 532, 534, 537, 560, 565, 567, 569, 571, 578, 586, 593, 595, 597, 600, 602, 604, 605, 611, 612, 616, 619, 621, 623, 624, 626, 628, 630, 635, 642
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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. A345491, A345522, A345533, A345535, A345543, A345579, A345786.

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])

A345836 Numbers that are the sum of eight fourth powers in exactly four ways.

Original entry on oeis.org

2933, 2948, 3013, 3173, 3188, 3557, 4148, 4163, 4213, 4293, 4388, 4453, 4643, 4772, 4837, 4883, 5012, 5123, 5188, 5203, 5268, 5333, 5363, 5378, 5398, 5428, 5538, 5573, 5603, 5618, 5668, 5733, 5748, 5858, 5923, 6052, 6163, 6227, 6292, 6548, 6578, 6628, 6693
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345579 at term 10 because 4228 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4.

Examples

			2948 is a term because 2948 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

Links

Crossrefs

Cf. A345579, A345786, A345826, A345835, A345837, A345846, A346329.

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, 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])

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])

A345785 Numbers that are the sum of eight cubes in exactly three ways.

Original entry on oeis.org

223, 230, 237, 249, 263, 270, 275, 282, 284, 286, 289, 291, 293, 308, 310, 312, 319, 326, 345, 349, 354, 364, 371, 373, 375, 378, 380, 385, 386, 387, 389, 397, 399, 404, 406, 410, 412, 413, 415, 420, 423, 439, 441, 443, 446, 449, 452, 453, 459, 460, 465, 473
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345533 at term 5 because 256 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3.
Likely finite.

Examples

			230 is a term because 230 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Links

Crossrefs

Cf. A345533, A345775, A345784, A345786, A345795, A345835.

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 == 3])
    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])

A345796 Numbers that are the sum of nine cubes in exactly four ways.

Original entry on oeis.org

224, 257, 264, 283, 320, 348, 355, 372, 374, 376, 381, 383, 390, 400, 402, 407, 411, 414, 416, 442, 450, 453, 454, 461, 474, 476, 481, 486, 488, 500, 503, 509, 510, 514, 519, 528, 529, 537, 542, 543, 544, 545, 548, 550, 552, 554, 555, 557, 564, 572, 573, 574
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345543 at term 17 because 409 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.
Likely finite.

Examples

			257 is a term because 257 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3.

Links

Crossrefs

Cf. A345543, A345786, A345795, A345797, A345806, A345846.

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, 9):
    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.

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