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.

A342686 Numbers that are the sum of five fifth powers in exactly two ways.

Original entry on oeis.org

4097, 51446, 51477, 51688, 52469, 54570, 59221, 68252, 68905, 84213, 110494, 131104, 151445, 212496, 300277, 325174, 325713, 355114, 422135, 422738, 589269, 637418, 794434, 810820, 876734, 876765, 876976, 877757, 879858, 884509, 893540, 909501, 924912, 935782, 976733, 995571, 1037784, 1083457
Offset: 1

Views

Author

David Consiglio, Jr., May 18 2021

Keywords

Comments

This sequence differs from A342685:
13124675 = 1^5 + 9^5 + 10^5 + 20^5 + 25^5
= 2^5 + 5^5 + 12^5 + 23^5 + 23^5
= 16^5 + 19^5 + 20^5 + 20^5 + 20^5,
so 13124675 is in A342685, but is not in this sequence.

Examples

			51477 = 2^5 + 4^5 + 7^5 + 7^5 + 7^5
      = 2^5 + 5^5 + 6^5 + 6^5 + 8^5
so 51477 is a term of this sequence.
		

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

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])

A344244 Numbers that are the sum of five fourth powers in exactly three ways.

Original entry on oeis.org

4225, 6610, 6850, 9170, 9235, 9490, 11299, 12929, 14209, 14690, 14755, 14770, 15314, 16579, 16594, 16659, 16834, 17203, 17235, 17315, 17859, 17874, 17939, 18785, 18850, 18979, 19154, 19700, 19715, 20674, 21250, 21330, 21364, 21410, 21954, 23139, 23795, 24754, 25810, 26578, 28610, 28930, 29330, 29699
Offset: 1

Views

Author

David Consiglio, Jr., May 12 2021

Keywords

Comments

Differs from A344243 at term 31 because 20995 = 1^4 + 1^4 + 1^4 + 4^4 + 12^4 = 2^4 + 3^4 + 3^4 + 3^4 + 12^4 = 2^4 + 6^4 + 9^4 + 9^4 + 9^4 = 4^4 + 6^4 + 7^4 + 7^4 + 11^4

Examples

			6850 is a member of this sequence because 6850 =  = 1^4 + 2^4 + 2^4 + 4^4 + 9^4 = 2^4 + 3^4 + 4^4 + 7^4 + 8^4 = 3^4 + 3^4 + 6^4 + 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(1,50)]
    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])

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])

A344518 Numbers that are the sum of five positive fifth powers in four or more ways.

Original entry on oeis.org

287618651, 1386406515, 1763135232, 2494769760, 2619898293, 3096064443, 3291315732, 3749564512, 4045994624, 5142310350, 5183605813, 5658934676, 5880926107, 7205217018, 7401155424, 7691215599, 8429499101, 8926086432, 9006349824, 9051501568, 9203796832
Offset: 1

Views

Author

David Consiglio, Jr., May 21 2021

Keywords

Examples

			287618651 is a term because
287618651 =  8^5 + 21^5 + 27^5 + 27^5 + 48^5
          =  9^5 + 13^5 + 26^5 + 37^5 + 46^5
          = 11^5 + 12^5 + 23^5 + 41^5 + 44^5
          = 11^5 + 20^5 + 22^5 + 30^5 + 48^5.
[Corrected by _Patrick De Geest_, Dec 28 2024]
		

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 >= 4])
    for x in range(len(rets)):
        print(rets[x])

A344519 Numbers that are the sum of five positive fifth powers in exactly four ways.

Original entry on oeis.org

287618651, 1386406515, 1763135232, 2494769760, 2619898293, 3096064443, 3291315732, 3749564512, 4045994624, 5142310350, 5183605813, 5658934676, 5880926107, 7205217018, 7401155424, 7691215599, 8429499101, 8926086432, 9051501568, 9203796832, 9254212901
Offset: 1

Views

Author

David Consiglio, Jr., May 21 2021

Keywords

Comments

Differs from A344518 at term 20 because
9006349824 = 8^5 + 34^5 + 62^5 + 68^5 + 92^5
= 8^5 + 41^5 + 47^5 + 79^5 + 89^5
= 12^5 + 18^5 + 72^5 + 78^5 + 84^5
= 21^5 + 34^5 + 43^5 + 74^5 + 92^5
= 24^5 + 42^5 + 48^5 + 54^5 + 96^5.

Examples

			287618651 is a term because
287618651 =  8^5 + 21^5 + 27^5 + 27^5 + 48^5
          =  9^5 + 13^5 + 26^5 + 37^5 + 46^5
          = 11^5 + 12^5 + 23^5 + 41^5 + 44^5
          = 11^5 + 20^5 + 22^5 + 30^5 + 48^5.
[Corrected by _Patrick De Geest_, Dec 28 2024]
		

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 == 4])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-6 of 6 results.