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.

A342687 Numbers that are the sum of five fifth powers in three or more ways.

Original entry on oeis.org

13124675, 28055699, 50043937, 52679923, 53069024, 55097976, 57936559, 60484744, 62260463, 62445305, 70211956, 73133026, 79401728, 80368962, 84766210, 88512249, 93288865, 98824300, 106993391, 113055482, 117173891, 120968132, 123383875, 126416258, 131106051, 131529588, 132022925
Offset: 1

Views

Author

David Consiglio, Jr., May 18 2021

Keywords

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, 500)]
    for pos in cwr(power_terms, 5):
        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])

A345560 Numbers that are the sum of six fourth powers in three or more ways.

Original entry on oeis.org

2676, 2851, 2916, 4131, 4226, 4241, 4306, 4371, 4481, 4850, 5346, 5411, 5521, 5586, 5651, 6561, 6611, 6626, 6691, 6756, 6771, 6801, 6821, 6836, 6851, 6866, 6931, 7106, 7235, 7475, 7491, 7666, 7841, 7906, 7971, 8146, 8211, 8321, 8386, 8451, 8531, 8706, 9011
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2851 is a term because 2851 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 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, 6):
        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])

A346358 Numbers that are the sum of six fifth powers in exactly three ways.

Original entry on oeis.org

696467, 893572, 1100264, 1109295, 1165727, 1711776, 2007401, 2025309, 2221767, 2801812, 3047519, 3310494, 3360608, 3550866, 3559556, 3576120, 3807122, 3907101, 4055922, 4093540, 4096114, 4104067, 4123363, 4135578, 4155107, 4195571, 4222339, 4326784, 4417112
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345604 at term 105 because 12047994 = 7^5 + 9^5 + 12^5 + 14^5 + 17^5 + 25^5 = 5^5 + 10^5 + 13^5 + 15^5 + 16^5 + 25^5 = 1^5 + 1^5 + 3^5 + 4^5 + 21^5 + 24^5 = 4^5 + 6^5 + 15^5 + 15^5 + 21^5 + 23^5.

Examples

			696467 is a term because 696467 = 1^5 + 6^5 + 8^5 + 9^5 + 9^5 + 14^5 = 3^5 + 3^5 + 7^5 + 9^5 + 12^5 + 13^5 = 4^5 + 4^5 + 4^5 + 11^5 + 11^5 + 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, 6):
        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])

A345507 Numbers that are the sum of six fifth powers in two or more ways.

Original entry on oeis.org

4098, 4129, 4340, 5121, 7222, 11873, 20904, 36865, 51447, 51478, 51509, 51689, 51720, 51931, 52470, 52501, 52712, 53493, 54571, 54602, 54813, 55594, 57695, 59222, 59253, 59464, 60245, 62346, 63146, 66997, 67586, 68253, 68284, 68495, 68906, 68937, 69148, 69276
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4129 is a term because 4129 = 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 = 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, 6):
        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])

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])

A345718 Numbers that are the sum of six fifth powers in four or more ways.

Original entry on oeis.org

12047994, 20646208, 21017489, 21300963, 21741819, 24993485, 27669050, 28576064, 30193856, 30785920, 35480456, 35735194, 36082750, 37303264, 39035975, 46814942, 47963291, 50047062, 50724345, 52987561, 53076800, 53606848, 54827300, 55101101, 56766906
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			20646208 is a term because 20646208 = 2^5 + 12^5 + 12^5 + 16^5 + 18^5 + 28^5 = 3^5 + 4^5 + 4^5 + 8^5 + 10^5 + 29^5 = 6^5 + 6^5 + 12^5 + 14^5 + 24^5 + 26^5 = 7^5 + 7^5 + 8^5 + 16^5 + 25^5 + 25^5.
		

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, 6):
        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])
Showing 1-6 of 6 results.