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.

A345834 Numbers that are the sum of eight fourth powers in exactly two ways.

Original entry on oeis.org

263, 278, 293, 308, 323, 343, 358, 373, 388, 423, 438, 453, 503, 533, 548, 563, 583, 598, 613, 628, 678, 693, 758, 773, 788, 803, 853, 868, 887, 902, 917, 932, 933, 967, 982, 997, 1028, 1043, 1047, 1062, 1108, 1127, 1142, 1157, 1172, 1222, 1237, 1283, 1302
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345577 at term 14 because 518 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

Examples

			278 is a term because 278 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.
		

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 == 2])
        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])

A346279 Numbers that are the sum of seven fifth powers in exactly two ways.

Original entry on oeis.org

4099, 4130, 4161, 4341, 4372, 4583, 5122, 5153, 5364, 6145, 7223, 7254, 7465, 8246, 10347, 11874, 11905, 12116, 12897, 14998, 19649, 20905, 20936, 21147, 21928, 24029, 28680, 36866, 36897, 37108, 37711, 37889, 39990, 40138, 44641, 51393, 51448, 51479, 51510
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345605 at term 156 because 84457 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 9^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5.

Examples

			4099 is a term because 4099 = 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.
		

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 == 2])
        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])

A346337 Numbers that are the sum of nine fifth powers in exactly two ways.

Original entry on oeis.org

4101, 4132, 4163, 4194, 4225, 4343, 4374, 4405, 4436, 4585, 4616, 4647, 4827, 4858, 5069, 5124, 5155, 5186, 5217, 5366, 5397, 5428, 5608, 5639, 5850, 6147, 6178, 6209, 6389, 6420, 6631, 7170, 7201, 7225, 7256, 7287, 7318, 7412, 7467, 7498, 7529, 7709, 7740
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345619 at term 306 because 52418 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.

Examples

			4101 is a term because 4101 = 1^5 + 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.
		

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 == 2])
        for x in range(len(rets)):
            print(rets[x])

A346326 Numbers that are the sum of eight fifth powers in exactly one way.

Original entry on oeis.org

8, 39, 70, 101, 132, 163, 194, 225, 250, 256, 281, 312, 343, 374, 405, 436, 467, 492, 523, 554, 585, 616, 647, 678, 734, 765, 796, 827, 858, 889, 976, 1007, 1031, 1038, 1062, 1069, 1093, 1100, 1124, 1155, 1186, 1217, 1218, 1248, 1249, 1273, 1280, 1304, 1311
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A003353 at term 156 because 4100 = 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.

Examples

			8 is a term because 8 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5.
		

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