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 1-7 of 7 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.
		

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, 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])

A345580 Numbers that are the sum of eight fourth powers in five or more ways.

Original entry on oeis.org

4228, 4403, 4468, 5443, 5508, 5683, 6613, 6643, 6658, 6708, 6723, 6773, 6788, 6838, 6853, 6868, 6883, 6898, 6948, 6963, 7013, 7028, 7093, 7138, 7203, 7267, 7268, 7332, 7397, 7478, 7507, 7572, 7588, 7828, 7858, 7923, 7938, 7988, 8003, 8068, 8113, 8133, 8178
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4403 is a term because 4403 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 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, 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])

A345588 Numbers that are the sum of nine fourth powers in four or more ways.

Original entry on oeis.org

2854, 2919, 2934, 2949, 2964, 3014, 3029, 3094, 3159, 3174, 3189, 3204, 3254, 3269, 3429, 3444, 3558, 3573, 3638, 3798, 3813, 3974, 4034, 4134, 4149, 4164, 4179, 4182, 4209, 4214, 4229, 4244, 4274, 4294, 4309, 4374, 4389, 4404, 4419, 4439, 4454, 4469, 4484
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2919 is a term because 2919 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 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, 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])

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])

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.
		

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 == 5])
        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])

A345622 Numbers that are the sum of nine fifth powers in five or more ways.

Original entry on oeis.org

392063, 392274, 406559, 458875, 519237, 538291, 607947, 663871, 672024, 672055, 672266, 672297, 673586, 673797, 674578, 675390, 680041, 681330, 704582, 704822, 714299, 730260, 732603, 763027, 763324, 765873, 766417, 777820, 780099, 814082, 820887, 825678
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			392274 is a term because 392274 = 1^5 + 1^5 + 4^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 1^5 + 3^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 2^5 + 3^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 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, 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])
Showing 1-7 of 7 results.