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.

A345566 Numbers that are the sum of six fourth powers in nine or more ways.

Original entry on oeis.org

88595, 122915, 132546, 134931, 144835, 146450, 151556, 161475, 162355, 162755, 170275, 171555, 171795, 172036, 172835, 173075, 177380, 177716, 180770, 183540, 183620, 184835, 185315, 185555, 187700, 187715, 190100, 190211, 193635, 195380, 195780, 196435
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			122915 is a term because 122915 = 1^4 + 3^4 + 6^4 + 9^4 + 10^4 + 18^4 = 1^4 + 4^4 + 7^4 + 8^4 + 15^4 + 16^4 = 1^4 + 7^4 + 9^4 + 10^4 + 14^4 + 16^4 = 2^4 + 3^4 + 4^4 + 5^4 + 14^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 11^4 + 18^4 = 2^4 + 9^4 + 9^4 + 12^4 + 14^4 + 15^4 = 3^4 + 5^4 + 6^4 + 6^4 + 11^4 + 18^4 = 3^4 + 8^4 + 10^4 + 11^4 + 13^4 + 16^4 = 5^4 + 6^4 + 7^4 + 11^4 + 14^4 + 16^4 = 8^4 + 8^4 + 9^4 + 10^4 + 11^4 + 17^4.
		

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 >= 9])
        for x in range(len(rets)):
            print(rets[x])

A345820 Numbers that are the sum of six fourth powers in exactly eight ways.

Original entry on oeis.org

58035, 59780, 87746, 96195, 96450, 102371, 106451, 106515, 108035, 108275, 108290, 108771, 112370, 112931, 115251, 122835, 122850, 124691, 125971, 133395, 133571, 133586, 134675, 136931, 138275, 138595, 143650, 144755, 145826, 147491, 148820, 149571, 150115
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345565 at term 4 because 88595 = 1^4 + 4^4 + 5^4 + 12^4 + 13^4 + 14^4 = 1^4 + 6^4 + 6^4 + 11^4 + 12^4 + 15^4 = 1^4 + 7^4 + 8^4 + 9^4 + 10^4 + 16^4 = 2^4 + 8^4 + 9^4 + 9^4 + 12^4 + 15^4 = 2^4 + 10^4 + 11^4 + 11^4 + 12^4 + 13^4 = 4^4 + 6^4 + 6^4 + 9^4 + 13^4 + 15^4 = 5^4 + 6^4 + 7^4 + 8^4 + 11^4 + 16^4 = 7^4 + 7^4 + 10^4 + 11^4 + 12^4 + 14^4.

Examples

			59780 is a term because 59780 = 1^4 + 1^4 + 1^4 + 5^4 + 12^4 + 14^4 = 1^4 + 1^4 + 6^4 + 6^4 + 9^4 + 15^4 = 1^4 + 2^4 + 9^4 + 10^4 + 11^4 + 13^4 = 1^4 + 4^4 + 7^4 + 7^4 + 8^4 + 15^4 = 1^4 + 7^4 + 7^4 + 9^4 + 10^4 + 14^4 = 2^4 + 5^4 + 6^4 + 11^4 + 11^4 + 13^4 = 3^4 + 7^4 + 8^4 + 10^4 + 11^4 + 13^4 = 5^4 + 6^4 + 7^4 + 7^4 + 11^4 + 14^4.
		

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 == 8])
        for x in range(len(rets)):
            print(rets[x])

A345831 Numbers that are the sum of seven fourth powers in exactly nine ways.

Original entry on oeis.org

19491, 21267, 21332, 23652, 35427, 36052, 37812, 38067, 39891, 40356, 41732, 41747, 43267, 43876, 43891, 43956, 44131, 44196, 44532, 44612, 45156, 45171, 45411, 45651, 45652, 45891, 46276, 46451, 46516, 47427, 48036, 48052, 48532, 48707, 49747, 49956, 49987
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345575 at term 5 because 31251 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 10^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 10^4 + 11^4 = 1^4 + 4^4 + 4^4 + 4^4 + 5^4 + 6^4 + 13^4 = 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 5^4 + 6^4 + 11^4 + 11^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 10^4 + 11^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 10^4 + 12^4 = 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 10^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 + 11^4 = 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 11^4.

Examples

			21267 is a term because 21267 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 2^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 + 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, 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 == 9])
        for x in range(len(rets)):
            print(rets[x])

A341892 Numbers that are the sum of five fourth powers in exactly nine ways.

Original entry on oeis.org

619090, 775714, 1100579, 1179379, 1186834, 1243699, 1357315, 1367539, 1373859, 1422595, 1431234, 1436419, 1511299, 1536019, 1699234, 1734899, 1839874, 1858594, 1880850, 1950355, 1951650, 1978915, 2044819, 2052899, 2069955, 2085139, 2101779, 2119459, 2133234
Offset: 1

Views

Author

David Consiglio, Jr., Jun 04 2021

Keywords

Comments

Differs from A341781 at term 3 because
954979 = 1^4 + 2^4 + 11^4 + 19^4 + 30^4
= 1^4 + 7^4 + 18^4 + 25^4 + 26^4
= 3^4 + 8^4 + 17^4 + 20^4 + 29^4
= 4^4 + 8^4 + 13^4 + 25^4 + 27^4
= 4^4 + 9^4 + 10^4 + 11^4 + 31^4
= 6^4 + 6^4 + 15^4 + 21^4 + 29^4
= 7^4 + 10^4 + 18^4 + 19^4 + 29^4
= 11^4 + 11^4 + 20^4 + 22^4 + 27^4
= 16^4 + 17^4 + 17^4 + 24^4 + 25^4
= 18^4 + 19^4 + 20^4 + 23^4 + 23^4.

Examples

			619090 =  1^4 +  2^4 + 18^4 + 22^4 + 23^4
       =  1^4 +  3^4 +  4^4 +  8^4 + 28^4
       =  1^4 + 11^4 + 14^4 + 22^4 + 24^4
       =  2^4 +  2^4 +  8^4 + 17^4 + 27^4
       =  2^4 + 13^4 + 13^4 + 18^4 + 26^4
       =  3^4 +  6^4 + 12^4 + 16^4 + 27^4
       =  4^4 + 12^4 + 14^4 + 23^4 + 23^4
       =  9^4 + 12^4 + 16^4 + 21^4 + 24^4
       = 14^4 + 16^4 + 18^4 + 19^4 + 23^4
so 619090 is a term.
		

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, 5):
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 9])
    for x in range(len(rets)):
        print(rets[x])

A345771 Numbers that are the sum of six cubes in exactly nine ways.

Original entry on oeis.org

2438, 2457, 2494, 2555, 2593, 2709, 2772, 2889, 2942, 2980, 3033, 3043, 3096, 3160, 3195, 3241, 3250, 3257, 3276, 3402, 3427, 3437, 3467, 3556, 3582, 3592, 3608, 3609, 3617, 3672, 3735, 3825, 3850, 3852, 3871, 3924, 3934, 3962, 3976, 3979, 3996, 3997, 4006
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345518 at term 14 because 3104 = 1^3 + 2^3 + 7^3 + 8^3 + 8^3 + 12^3 = 1^3 + 5^3 + 5^3 + 5^3 + 10^3 + 12^3 = 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 14^3 = 2^3 + 3^3 + 4^3 + 7^3 + 11^3 + 11^3 = 2^3 + 3^3 + 5^3 + 6^3 + 10^3 + 12^3 = 2^3 + 7^3 + 8^3 + 8^3 + 9^3 + 10^3 = 3^3 + 3^3 + 5^3 + 6^3 + 8^3 + 13^3 = 4^3 + 5^3 + 7^3 + 8^3 + 9^3 + 11^3 = 5^3 + 5^3 + 5^3 + 9^3 + 10^3 + 10^3 = 5^3 + 6^3 + 6^3 + 6^3 + 10^3 + 11^3 = 6^3 + 6^3 + 6^3 + 6^3 + 8^3 + 12^3.

Examples

			2457 is a term because 2457 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 12^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 12^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 11^3 = 1^3 + 5^3 + 5^3 + 7^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 11^3 = 2^3 + 3^3 + 3^3 + 3^3 + 9^3 + 10^3 = 2^3 + 5^3 + 5^3 + 6^3 + 6^3 + 10^3 = 3^3 + 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 3^3 + 4^3 + 7^3 + 8^3 + 9^3.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**3 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 == 9])
        for x in range(len(rets)):
            print(rets[x])

A345822 Numbers that are the sum of six fourth powers in exactly ten ways.

Original entry on oeis.org

122915, 151556, 161475, 162755, 173075, 183620, 185315, 199106, 199940, 201875, 202275, 204275, 204340, 204595, 206115, 207395, 209795, 211075, 213731, 217826, 217891, 218515, 221250, 223955, 224180, 225875, 226595, 227186, 228035, 236195, 237796, 237890
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345567 at term 8 because 197795 = 1^4 + 2^4 + 5^4 + 6^4 + 16^4 + 19^4 = 1^4 + 2^4 + 7^4 + 11^4 + 12^4 + 20^4 = 1^4 + 2^4 + 10^4 + 12^4 + 17^4 + 17^4 = 2^4 + 4^4 + 7^4 + 9^4 + 13^4 + 20^4 = 2^4 + 11^4 + 13^4 + 14^4 + 15^4 + 16^4 = 3^4 + 6^4 + 6^4 + 9^4 + 13^4 + 20^4 = 3^4 + 6^4 + 7^4 + 14^4 + 15^4 + 18^4 = 4^4 + 9^4 + 11^4 + 12^4 + 15^4 + 18^4 = 7^4 + 7^4 + 14^4 + 14^4 + 15^4 + 16^4.

Examples

			151556 is a term because 151556 = 1^4 + 2^4 + 2^4 + 9^4 + 11^4 + 19^4 = 1^4 + 2^4 + 3^4 + 7^4 + 16^4 + 17^4 = 1^4 + 8^4 + 11^4 + 12^4 + 13^4 + 17^4 = 2^4 + 3^4 + 7^4 + 8^4 + 11^4 + 19^4 = 3^4 + 3^4 + 3^4 + 4^4 + 12^4 + 19^4 = 3^4 + 4^4 + 11^4 + 11^4 + 14^4 + 17^4 = 3^4 + 4^4 + 13^4 + 13^4 + 13^4 + 16^4 = 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 19^4 = 4^4 + 7^4 + 11^4 + 11^4 + 11^4 + 18^4 = 4^4 + 8^4 + 9^4 + 13^4 + 13^4 + 17^4.
		

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 == 10])
        for x in range(len(rets)):
            print(rets[x])

A346364 Numbers that are the sum of six fifth powers in exactly nine ways.

Original entry on oeis.org

9085584992, 16933805856, 37377003050, 39254220544, 41066625600, 41485873792, 42149876800, 43828403850, 44180505600, 45902654525, 48588434400, 52005184992, 53536896864, 54156285568, 56229189632, 57088402525, 59954496800, 63432407850, 66188522400, 66507304800
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

This sequence differs from A345723:
55302546200 = 34^5 + 38^5 + 50^5 + 57^5 + 95^5 + 136^5
= 23^5 + 49^5 + 61^5 + 69^5 + 107^5 + 131^5
= 24^5 + 37^5 + 63^5 + 81^5 + 104^5 + 131^5
= 21^5 + 35^5 + 60^5 + 94^5 + 100^5 + 130^5
= 57^5 + 60^5 + 71^5 + 75^5 + 109^5 + 128^5
= 19^5 + 37^5 + 56^5 + 96^5 + 104^5 + 128^5
= 35^5 + 41^5 + 53^5 + 69^5 + 115^5 + 127^5
= 16^5 + 49^5 + 53^5 + 83^5 + 112^5 + 127^5
= 35^5 + 37^5 + 40^5 + 88^5 + 119^5 + 121^5
= 11^5 + 24^5 + 71^5 + 104^5 + 109^5 + 121^5,
so 55302546200 is in A345723, but is not in this sequence.

Examples

			9085584992 = 24^5 + 38^5 + 42^5 + 48^5 + 54^5 + 96^5
           = 21^5 + 34^5 + 38^5 + 43^5 + 74^5 + 92^5
           =  8^5 + 34^5 + 38^5 + 62^5 + 68^5 + 92^5
           = 18^5 + 18^5 + 44^5 + 64^5 + 66^5 + 92^5
           = 13^5 + 18^5 + 51^5 + 64^5 + 64^5 + 92^5
           =  8^5 + 38^5 + 41^5 + 47^5 + 79^5 + 89^5
           =  5^5 + 23^5 + 29^5 + 45^5 + 85^5 + 85^5
           =  8^5 + 23^5 + 41^5 + 64^5 + 82^5 + 84^5
           = 12^5 + 18^5 + 38^5 + 72^5 + 78^5 + 84^5,
so 9085584992 is a term.
		

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 == 9])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-7 of 7 results.