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.

A344358 Numbers that are the sum of five fourth powers in five or more ways.

Original entry on oeis.org

59779, 67859, 93394, 108274, 112850, 136915, 142354, 151300, 151475, 161459, 168979, 181219, 183539, 183604, 185299, 187699, 189394, 193379, 195394, 197779, 199090, 199474, 200979, 201874, 202979, 203299, 205859, 211059, 211330, 212419, 213730, 217154, 217810, 217890, 221779, 223090, 223155
Offset: 1

Views

Author

David Consiglio, Jr., May 15 2021

Keywords

Examples

			93394 is a term of this sequence because 93394 = 1^4 + 4^4 + 8^4 + 14^4 + 15^4 = 1^4 + 6^4 + 12^4 + 12^4 + 15^4 = 1^4 + 9^4 + 10^4 + 14^4 + 14^4 = 5^4 + 6^4 + 11^4 + 14^4 + 14^4 = 5^4 + 7^4 + 8^4 + 12^4 + 16^4.
		

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])

A342688 Numbers that are the sum of five positive fifth powers in exactly three 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

Comments

Differs from A342687:
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.
So 287618651 is a term of A342687 but not a term of this sequence.
[Corrected by Patrick De Geest, Dec 28 2024]

Examples

			50043937 =  6^5 + 16^5 + 18^5 + 24^5 + 33^5
         =  7^5 + 13^5 + 21^5 + 23^5 + 33^5
         = 11^5 + 13^5 + 13^5 + 29^5 + 31^5
so 50043937 is a term of this sequence.
[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 == 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])

A345719 Numbers that are the sum of six fifth powers in five or more ways.

Original entry on oeis.org

54827300, 74115800, 74883600, 75609125, 113088250, 120274275, 166078869, 169692136, 174781858, 178736448, 182341225, 185558208, 194939538, 203054589, 218814275, 235067008, 250989825, 251772882, 252721458, 255453233, 258124975, 274616694, 282859667
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			74115800 is a term because 74115800 = 1^5 + 4^5 + 21^5 + 21^5 + 29^5 + 34^5 = 1^5 + 8^5 + 14^5 + 23^5 + 32^5 + 32^5 = 4^5 + 11^5 + 13^5 + 22^5 + 24^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 24^5 + 36^5 = 6^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5.
		

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 >= 5])
        for x in range(len(rets)):
            print(rets[x])

A346257 Numbers that are the sum of five fifth powers in exactly five ways.

Original entry on oeis.org

9006349824, 65799210368, 67629776576, 181085909632, 188189635424, 295677350451, 467139768468, 471359089024, 656243139157, 691381929281, 797466940832, 854533526901, 874953049024, 891862586132, 953769598750, 1038549256768, 1092458681568, 1182658308657
Offset: 1

Views

Author

David Consiglio, Jr., Jul 11 2021

Keywords

Comments

Differs from 103 terms known at term 6 because 288203194368 = 48^5 + 84^5 + 96^5 + 108^5 + 192^5 = 16^5 + 99^5 + 103^5 + 121^5 + 189^5 = 42^5 + 68^5 + 86^5 + 148^5 + 184^5 = 16^5 + 68^5 + 124^5 + 136^5 + 184^5 = 16^5 + 82^5 + 94^5 + 158^5 + 178^5 = 24^5 + 36^5 + 144^5 + 156^5 + 168^5.

Examples

			9006349824 is a term because 9006349824 = 24^5 + 42^5 + 48^5 + 54^5 + 96^5 = 21^5 + 34^5 + 43^5 + 74^5 + 92^5 = 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.
		

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, 5):
        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])

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])

A345864 Numbers that are the sum of five fifth powers in six or more ways.

Original entry on oeis.org

288203194368, 2784485221024, 6022068333568, 9222502219776, 9670153077344, 10918228000032, 15441787364576
Offset: 1

Views

Author

David Consiglio, Jr., Jun 27 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, 1000)]
    for pos in cwr(power_terms, 5):
        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])
Showing 1-7 of 7 results.