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.

A345518 Numbers that are the sum of six cubes in nine or more ways.

Original entry on oeis.org

2438, 2457, 2494, 2555, 2593, 2709, 2772, 2889, 2942, 2980, 3033, 3043, 3096, 3104, 3160, 3195, 3215, 3222, 3241, 3250, 3257, 3267, 3276, 3313, 3339, 3374, 3402, 3427, 3430, 3437, 3465, 3467, 3491, 3493, 3528, 3547, 3556, 3582, 3584, 3592, 3608, 3609, 3617
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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])

A345565 Numbers that are the sum of six fourth powers in eight or more ways.

Original entry on oeis.org

58035, 59780, 87746, 88595, 96195, 96450, 102371, 106451, 106515, 108035, 108275, 108290, 108771, 112370, 112931, 115251, 122835, 122850, 122915, 124691, 125971, 132546, 133395, 133571, 133586, 134675, 134931, 136931, 138275, 138595, 143650, 144755, 144835
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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])

A345575 Numbers that are the sum of seven fourth powers in nine or more ways.

Original entry on oeis.org

19491, 21267, 21332, 23652, 31251, 35427, 36052, 37812, 38067, 39891, 40356, 41732, 41747, 43267, 43876, 43891, 43956, 44131, 44196, 44532, 44547, 44612, 45156, 45171, 45411, 45651, 45652, 45827, 45891, 45892, 45907, 46276, 46451, 46516, 47427, 47667, 47971
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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])

A345821 Numbers that are the sum of six fourth powers in exactly nine ways.

Original entry on oeis.org

88595, 132546, 134931, 144835, 146450, 162355, 170275, 171555, 171795, 172036, 172835, 177380, 177716, 180770, 183540, 184835, 185555, 187700, 187715, 190100, 190211, 193635, 195380, 195780, 196435, 197780, 199075, 199475, 199730, 199955, 202196, 202980
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345566 at term 2 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.

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])

A341891 Numbers that are the sum of five fourth powers in nine or more ways.

Original entry on oeis.org

619090, 775714, 954979, 1100579, 1179379, 1186834, 1205539, 1243699, 1357315, 1367539, 1373859, 1422595, 1431234, 1436419, 1511299, 1536019, 1574850, 1699234, 1713859, 1734899, 1801459, 1839874, 1858594, 1863859, 1877394, 1880850, 1882579, 1950355, 1951650
Offset: 1

Views

Author

David Consiglio, Jr., Jun 04 2021

Keywords

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])

A345567 Numbers that are the sum of six fourth powers in ten or more ways.

Original entry on oeis.org

122915, 151556, 161475, 162755, 173075, 183620, 185315, 197795, 199106, 199940, 201875, 201955, 202275, 204275, 204340, 204595, 206115, 207395, 209795, 211075, 212420, 213731, 217620, 217826, 217891, 218515, 221250, 223715, 223955, 224180, 224451, 225875
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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])

A345723 Numbers that are the sum of six fifth powers in nine or more ways.

Original entry on oeis.org

9085584992, 16933805856, 37377003050, 39254220544, 41066625600, 41485873792, 42149876800, 43828403850, 44180505600, 45902654525, 48588434400, 52005184992, 53536896864, 54156285568, 55302546200, 56229189632, 57088402525, 59954496800, 63432407850
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			16933805856 =  2^5 + 38^5 + 68^5 + 74^5 + 92^5 +  92^5
            =  2^5 + 54^5 + 58^5 + 64^5 + 92^5 +  96^5
            = 14^5 + 36^5 + 61^5 + 67^5 + 94^5 +  94^5
            = 15^5 + 49^5 + 52^5 + 60^5 + 94^5 +  96^5
            = 17^5 + 49^5 + 53^5 + 57^5 + 92^5 +  98^5
            = 29^5 + 36^5 + 42^5 + 72^5 + 88^5 +  99^5
            = 31^5 + 36^5 + 54^5 + 54^5 + 94^5 +  97^5
            = 34^5 + 34^5 + 46^5 + 72^5 + 76^5 + 104^5
            = 35^5 + 36^5 + 69^5 + 72^5 + 89^5 +  95^5
so 16933805856 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.