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.

A345578 Numbers that are the sum of eight fourth powers in three or more ways.

Original entry on oeis.org

518, 2678, 2693, 2708, 2738, 2758, 2773, 2838, 2853, 2868, 2883, 2918, 2933, 2948, 2998, 3013, 3078, 3108, 3123, 3173, 3188, 3253, 3302, 3317, 3363, 3382, 3428, 3477, 3492, 3542, 3557, 3622, 3732, 3778, 3797, 3893, 3926, 3953, 3973, 3988, 4018, 4053, 4101
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2678 is a term because 2678 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 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, 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])

A345606 Numbers that are the sum of seven fifth powers in three or more ways.

Original entry on oeis.org

84457, 166997, 324860, 326199, 358482, 359327, 391007, 391999, 408158, 455146, 455749, 486468, 502429, 572054, 595519, 614505, 622280, 648319, 671210, 672022, 696468, 696499, 696710, 697491, 699592, 704243, 713274, 729235, 755516, 796467, 857518, 877645
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			166997 is a term because 166997 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 = 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 10^5 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 10^5.
		

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, 7):
        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])

A345610 Numbers that are the sum of eight fifth powers in two or more ways.

Original entry on oeis.org

4100, 4131, 4162, 4193, 4342, 4373, 4404, 4584, 4615, 4826, 5123, 5154, 5185, 5365, 5396, 5607, 6146, 6177, 6388, 7169, 7224, 7255, 7286, 7466, 7497, 7708, 8247, 8278, 8489, 9270, 10348, 10379, 10590, 11371, 11875, 11906, 11937, 12117, 12148, 12359, 12898
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4131 is a term because 4131 = 1^5 + 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5.
		

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, 8):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 2])
        for x in range(len(rets)):
            print(rets[x])

A345612 Numbers that are the sum of eight fifth powers in four or more ways.

Original entry on oeis.org

391250, 392031, 455750, 519236, 604822, 622281, 672023, 672054, 672265, 673554, 697492, 703978, 707368, 730259, 763292, 857761, 893605, 893636, 893816, 893847, 894027, 894058, 894452, 894628, 896729, 897151, 901380, 903839, 909124, 909597, 910411, 911403
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			392031 is a term because 392031 = 1^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 2^5 + 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, 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])

A345620 Numbers that are the sum of nine fifth powers in three or more ways.

Original entry on oeis.org

52418, 52449, 52660, 53441, 54519, 54550, 54761, 55542, 55690, 57643, 60193, 62294, 69224, 69635, 69666, 69877, 70658, 70955, 70986, 71197, 71325, 71978, 72759, 73001, 74079, 76031, 77410, 78730, 84162, 84459, 84490, 84521, 84701, 84732, 84943, 85185, 85482
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			52449 is a term because 52449 = 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 = 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5.
		

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 >= 3])
        for x in range(len(rets)):
            print(rets[x])

A346328 Numbers that are the sum of eight fifth powers in exactly three ways.

Original entry on oeis.org

52417, 54518, 69634, 70954, 84458, 84489, 84700, 85481, 87582, 92233, 101264, 102890, 112574, 117225, 119326, 134473, 143264, 143442, 143506, 149781, 151448, 158719, 159465, 165634, 166998, 167029, 167196, 167240, 168021, 170122, 174773, 183804, 184457
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345611 at term 105 because 391250 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5.

Examples

			52417 is a term because 52417 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.
		

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, 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])
Showing 1-6 of 6 results.