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.

A345544 Numbers that are the sum of nine cubes in five or more ways.

Original entry on oeis.org

409, 413, 428, 435, 439, 446, 465, 472, 479, 491, 498, 502, 505, 507, 512, 517, 524, 526, 531, 533, 535, 538, 540, 559, 561, 563, 566, 568, 570, 576, 579, 580, 587, 594, 596, 598, 600, 601, 603, 605, 613, 615, 617, 620, 622, 624, 627, 629, 631, 632, 633, 635
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			413 is a term because 413 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.

Links

Crossrefs

Cf. A345502, A345535, A345543, A345545, A345553, A345589, A345797.

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])

A345847 Numbers that are the sum of nine fourth powers in exactly five ways.

Original entry on oeis.org

3189, 4149, 4229, 4244, 4309, 4374, 4404, 4419, 4549, 4659, 4724, 4853, 4899, 5028, 5093, 5139, 5189, 5204, 5269, 5284, 5349, 5379, 5414, 5509, 5574, 5619, 5634, 5654, 5684, 5749, 5814, 5829, 5939, 6068, 6133, 6179, 6308, 6419, 6564, 6594, 6614, 6644, 6709
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345589 at term 9 because 4469 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4.

Examples

			4149 is a term because 4149 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 = 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4.

Links

Crossrefs

Cf. A345589, A345797, A345837, A345846, A345848, A345857, A346340.

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, 9):
    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])

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])

A345798 Numbers that are the sum of nine cubes in exactly six ways.

Original entry on oeis.org

472, 498, 505, 507, 524, 596, 598, 605, 636, 643, 655, 661, 662, 669, 672, 676, 681, 688, 690, 692, 696, 706, 718, 722, 725, 728, 729, 731, 732, 737, 739, 742, 748, 749, 750, 751, 756, 765, 772, 782, 783, 785, 787, 788, 791, 793, 794, 800, 801, 802, 808, 810
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345545 at term 9 because 624 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3.
Likely finite.

Examples

			498 is a term because 498 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3.

Links

Crossrefs

Cf. A345545, A345788, A345797, A345799, A345808, A345848.

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

A345807 Numbers that are the sum of ten cubes in exactly five ways.

Original entry on oeis.org

288, 349, 382, 384, 401, 403, 408, 410, 414, 415, 417, 421, 429, 443, 454, 455, 462, 475, 482, 487, 496, 501, 520, 543, 544, 545, 546, 548, 555, 556, 558, 565, 566, 570, 573, 574, 575, 576, 578, 580, 582, 586, 587, 591, 592, 593, 594, 600, 609, 610, 611, 617
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345553 at term 14 because 436 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.
Likely finite.

Examples

			349 is a term because 349 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.

Links

Crossrefs

Cf. A345553, A345797, A345806, A345808, A345857.

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, 10):
    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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