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.

A345590 Numbers that are the sum of nine fourth powers in six or more ways.

Original entry on oeis.org

4469, 4484, 5444, 5459, 5524, 5589, 5699, 5764, 6629, 6659, 6674, 6694, 6724, 6739, 6789, 6804, 6854, 6869, 6884, 6899, 6914, 6934, 6949, 6964, 6979, 7014, 7029, 7044, 7094, 7109, 7154, 7219, 7269, 7284, 7334, 7348, 7349, 7413, 7459, 7478, 7494, 7523, 7524
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4484 is a term because 4484 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.
		

Crossrefs

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

A345554 Numbers that are the sum of ten cubes in six or more ways.

Original entry on oeis.org

436, 440, 447, 466, 473, 477, 480, 492, 499, 503, 506, 508, 510, 513, 515, 518, 525, 527, 529, 532, 534, 536, 538, 539, 541, 551, 553, 560, 562, 564, 567, 569, 571, 577, 581, 584, 588, 590, 595, 597, 599, 601, 602, 603, 604, 606, 607, 608, 613, 614, 616, 618
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			440 is a term because 440 = 1^3 + 1^3 + 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 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1^3 + 1^3 + 3^3 + 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 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 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 + 3^3 + 5^3.
		

Crossrefs

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

A345598 Numbers that are the sum of ten fourth powers in five or more ways.

Original entry on oeis.org

2935, 3110, 3175, 3190, 3205, 3270, 3445, 3814, 3940, 4150, 4165, 4180, 4195, 4215, 4230, 4245, 4260, 4290, 4310, 4325, 4375, 4390, 4405, 4420, 4435, 4455, 4470, 4485, 4500, 4550, 4565, 4615, 4630, 4660, 4675, 4695, 4725, 4740, 4774, 4805, 4854, 4869, 4870
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			3110 is a term because 3110 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.
		

Crossrefs

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, 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])

A345600 Numbers that are the sum of ten fourth powers in seven or more ways.

Original entry on oeis.org

4485, 5445, 5460, 5525, 5540, 5590, 5605, 5670, 5700, 5715, 5765, 5780, 5830, 5845, 6645, 6675, 6710, 6740, 6755, 6775, 6805, 6820, 6855, 6870, 6885, 6900, 6915, 6930, 6935, 6950, 6965, 6980, 6995, 7015, 7030, 7045, 7060, 7095, 7110, 7125, 7175, 7190, 7235
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			5445 is a term because 5445 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 = 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 8^4 = 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4.
		

Crossrefs

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, 10):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 7])
        for x in range(len(rets)):
            print(rets[x])

A345858 Numbers that are the sum of ten fourth powers in exactly six ways.

Original entry on oeis.org

3175, 4150, 4230, 4390, 4405, 4455, 4470, 4500, 4550, 4565, 4630, 4725, 4740, 4915, 4980, 5094, 5109, 5155, 5190, 5205, 5220, 5270, 5285, 5350, 5365, 5395, 5430, 5475, 5635, 5655, 5735, 5910, 5955, 6020, 6069, 6084, 6149, 6195, 6214, 6324, 6389, 6435, 6500
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345599 at term 8 because 4485 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4.

Examples

			4150 is a term because 4150 = 1^4 + 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 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 = 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 7^4.
		

Crossrefs

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, 10):
        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])

A345638 Numbers that are the sum of ten fifth powers in six or more ways.

Original entry on oeis.org

392095, 392306, 399839, 406802, 407583, 434676, 491643, 492063, 520261, 521106, 538323, 538534, 540927, 553325, 555098, 563526, 582089, 592398, 608190, 611072, 614196, 637833, 639903, 640715, 640895, 640926, 640957, 641106, 643671, 653523, 655327, 656616
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			392306 is a term because 392306 = 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5 = 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 1^5 + 2^5 + 3^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 2^5 + 2^5 + 3^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 3^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 2^5 + 2^5 + 3^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**5 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 >= 6])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.