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.

A345591 Numbers that are the sum of nine fourth powers in seven or more ways.

Original entry on oeis.org

6739, 6804, 6854, 6869, 6979, 7029, 7044, 7094, 7109, 7269, 7284, 7844, 7909, 7939, 8004, 8019, 8084, 8149, 8194, 8244, 8259, 8309, 8324, 8389, 8434, 8499, 8564, 8628, 8739, 8868, 8979, 9044, 9059, 9124, 9189, 9219, 9234, 9254, 9284, 9299, 9364, 9414, 9429
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6804 is a term because 6804 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 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 >= 7])
        for x in range(len(rets)):
            print(rets[x])

A345555 Numbers that are the sum of ten cubes in seven or more ways.

Original entry on oeis.org

440, 473, 499, 506, 525, 532, 534, 567, 571, 584, 588, 597, 599, 604, 606, 623, 625, 627, 630, 632, 637, 639, 640, 644, 651, 656, 658, 660, 662, 663, 665, 669, 670, 673, 677, 680, 682, 684, 688, 689, 691, 693, 695, 696, 697, 699, 701, 702, 704, 707, 708, 714
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			473 is a term because 473 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.
		

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 >= 7])
        for x in range(len(rets)):
            print(rets[x])

A345599 Numbers that are the sum of ten fourth powers in six or more ways.

Original entry on oeis.org

3175, 4150, 4230, 4390, 4405, 4455, 4470, 4485, 4500, 4550, 4565, 4630, 4725, 4740, 4915, 4980, 5094, 5109, 5155, 5190, 5205, 5220, 5270, 5285, 5350, 5365, 5395, 5430, 5445, 5460, 5475, 5525, 5540, 5590, 5605, 5635, 5655, 5670, 5700, 5715, 5735, 5765, 5780
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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])

A345601 Numbers that are the sum of ten fourth powers in eight or more ways.

Original entry on oeis.org

6675, 6740, 6755, 6805, 6820, 6870, 6885, 6950, 6995, 7015, 7030, 7045, 7060, 7095, 7110, 7125, 7270, 7285, 7300, 7350, 7365, 7429, 7494, 7525, 7540, 7590, 7605, 7750, 7780, 7845, 7860, 7925, 7955, 7990, 8005, 8020, 8035, 8085, 8100, 8150, 8165, 8195, 8215
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6740 is a term because 6740 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4.
		

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 >= 8])
        for x in range(len(rets)):
            print(rets[x])

A345859 Numbers that are the sum of ten fourth powers in exactly seven ways.

Original entry on oeis.org

4485, 5445, 5460, 5525, 5540, 5590, 5605, 5670, 5700, 5715, 5765, 5780, 5830, 5845, 6645, 6710, 6775, 6855, 6900, 6915, 6930, 6935, 6965, 6980, 7175, 7190, 7235, 7255, 7335, 7364, 7415, 7430, 7475, 7479, 7495, 7510, 7604, 7620, 7654, 7669, 7670, 7685, 7715
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345600 at term 16 because 6675 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4 = 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4.

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])

A345639 Numbers that are the sum of ten fifth powers in seven or more ways.

Original entry on oeis.org

555098, 674040, 683166, 707315, 763631, 777852, 778844, 780945, 783224, 893654, 896500, 897668, 920887, 926616, 927819, 928802, 936850, 937631, 944383, 945017, 952897, 953077, 953139, 953350, 953414, 955178, 963131, 975133, 979482, 984133, 985453, 985664
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			674040 is a term because 674040 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 7^5 + 9^5 + 10^5 + 12^5 + 12^5 = 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 = 1^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 = 1^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 = 2^5 + 2^5 + 2^5 + 2^5 + 4^5 + 6^5 + 8^5 + 10^5 + 11^5 + 13^5 = 2^5 + 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 6^5 + 9^5 + 9^5 + 14^5 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 7^5 + 8^5 + 12^5 + 13^5.
		

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 >= 7])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.