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.

A345527 Numbers that are the sum of seven cubes in nine or more ways.

Original entry on oeis.org

1496, 1648, 1704, 1711, 1720, 1737, 1772, 1774, 1781, 1800, 1802, 1835, 1837, 1844, 1863, 1882, 1889, 1891, 1893, 1898, 1900, 1907, 1912, 1919, 1926, 1938, 1945, 1952, 1954, 1961, 1963, 1982, 1989, 1996, 2000, 2008, 2012, 2015, 2019, 2026, 2045, 2052, 2053
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1648 is a term because 1648 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 5^3 + 6^3 + 6^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 + 8^3 = 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 7^3 + 7^3.
		

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, 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])

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])

A345574 Numbers that are the sum of seven fourth powers in eight or more ways.

Original entry on oeis.org

19491, 21252, 21267, 21332, 21507, 21636, 21876, 23652, 25347, 30372, 31251, 31412, 31652, 32116, 32356, 33811, 33907, 35427, 35637, 35652, 35892, 36052, 36261, 37812, 37827, 38052, 38067, 38596, 38676, 39267, 39347, 39891, 39971, 39972, 40212, 40356, 40452
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			21252 is a term because 21252 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 1^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 7^4 + 8^4 + 9^4 + 9^4.
		

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

A345584 Numbers that are the sum of eight fourth powers in nine or more ways.

Original entry on oeis.org

15427, 16692, 17348, 17493, 17972, 17987, 18052, 18227, 19267, 19412, 19492, 19507, 19572, 19747, 19748, 20116, 20787, 20852, 21268, 21283, 21333, 21348, 21413, 21443, 21493, 21508, 21523, 21588, 21637, 21652, 21653, 21827, 21877, 21892, 21957, 21972, 22037
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			16692 is a term because 16692 = 1^4 + 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 + 10^4 = 1^4 + 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 11^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 10^4 = 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 1^4 + 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 5^4 + 7^4 + 9^4 + 9^4 = 2^4 + 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 9^4 + 9^4.
		

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, 8):
        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])

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])

A345576 Numbers that are the sum of seven fourth powers in ten or more ways.

Original entry on oeis.org

31251, 44547, 45827, 45892, 45907, 47667, 47971, 48292, 49572, 49812, 50052, 51092, 52372, 53316, 53476, 54531, 54596, 54756, 54996, 57411, 58036, 58116, 58276, 58516, 58660, 58756, 59331, 59781, 59796, 59811, 59827, 59861, 59876, 59892, 60036, 60371, 60436
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			44547 is a term because 44547 = 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 11^4 + 13^4 = 1^4 + 2^4 + 2^4 + 6^4 + 7^4 + 7^4 + 14^4 = 1^4 + 2^4 + 6^4 + 6^4 + 9^4 + 11^4 + 12^4 = 1^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 13^4 = 2^4 + 2^4 + 8^4 + 9^4 + 9^4 + 9^4 + 12^4 = 2^4 + 4^4 + 6^4 + 6^4 + 9^4 + 9^4 + 13^4 = 2^4 + 4^4 + 7^4 + 7^4 + 8^4 + 11^4 + 12^4 = 3^4 + 3^4 + 4^4 + 4^4 + 7^4 + 12^4 + 12^4 = 3^4 + 6^4 + 6^4 + 7^4 + 8^4 + 11^4 + 12^4 = 4^4 + 4^4 + 8^4 + 8^4 + 9^4 + 11^4 + 11^4.
		

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

A345631 Numbers that are the sum of seven fifth powers in nine or more ways.

Original entry on oeis.org

110276376, 124732805, 127808693, 130298618, 134581976, 188116743, 189642309, 202274051, 202686274, 203343582, 219063107, 230909843, 233137574, 233549568, 234250752, 235438301, 244250335, 251138524, 252277376, 253480833, 254017026, 254380543
Offset: 1

Views

Author

David Consiglio, Jr., Jun 22 2021

Keywords

Examples

			124732805 is a term because 124732805 = 3^5 + 18^5 + 22^5 + 22^5 + 24^5 + 27^5 + 39^5 = 4^5 + 15^5 + 17^5 + 21^5 + 29^5 + 34^5 + 35^5 = 5^5 + 14^5 + 17^5 + 24^5 + 25^5 + 35^5 + 35^5 = 7^5 + 8^5 + 17^5 + 26^5 + 29^5 + 34^5 + 34^5 = 7^5 + 10^5 + 12^5 + 31^5 + 31^5 + 32^5 + 32^5 = 7^5 + 12^5 + 23^5 + 24^5 + 24^5 + 26^5 + 39^5 = 7^5 + 14^5 + 22^5 + 22^5 + 23^5 + 28^5 + 39^5 = 16^5 + 25^5 + 25^5 + 28^5 + 28^5 + 28^5 + 35^5 = 20^5 + 24^5 + 24^5 + 25^5 + 25^5 + 32^5 + 35^5.
		

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