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.

A345591 Numbers that are the sum of nine fourth powers in seven or more ways.

Original entry on oeis.org

6739, 6804, 6854, 6869, 6979, 7029, 7044, 7094, 7109, 7269, 7284, 7844, 7909, 7939, 8004, 8019, 8084, 8149, 8194, 8244, 8259, 8309, 8324, 8389, 8434, 8499, 8564, 8628, 8739, 8868, 8979, 9044, 9059, 9124, 9189, 9219, 9234, 9254, 9284, 9299, 9364, 9414, 9429
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6804 is a term because 6804 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.
		

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, 9):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 7])
        for x in range(len(rets)):
            print(rets[x])

A345839 Numbers that are the sum of eight fourth powers in exactly seven ways.

Original entry on oeis.org

8003, 8243, 9043, 9218, 9283, 9523, 10372, 10803, 10868, 10948, 11043, 11412, 11557, 11587, 12083, 12692, 12932, 13188, 13333, 13508, 13972, 14147, 14387, 14883, 14933, 14948, 14963, 15013, 15028, 15093, 15173, 15268, 15317, 15332, 15397, 15412, 15413, 15492
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345582 at term 19 because 13268 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 8^4 + 9^4 = 1^4 + 1^4 + 2^4 + 4^4 + 7^4 + 7^4 + 8^4 + 8^4 = 1^4 + 1^4 + 3^4 + 6^4 + 6^4 + 7^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 9^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 9^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 9^4 = 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 10^4.

Examples

			8243 is a term because 8243 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 8^4 = 2^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 3^4 + 4^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4.
		

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 == 7])
        for x in range(len(rets)):
            print(rets[x])

A345848 Numbers that are the sum of nine fourth powers in exactly six ways.

Original entry on oeis.org

4469, 4484, 5444, 5459, 5524, 5589, 5699, 5764, 6629, 6659, 6674, 6694, 6724, 6789, 6884, 6899, 6914, 6934, 6949, 6964, 7014, 7154, 7219, 7334, 7348, 7349, 7413, 7459, 7478, 7494, 7523, 7524, 7588, 7589, 7604, 7653, 7669, 7734, 7779, 7874, 7954, 7989, 8069
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345590 at term 14 because 6739 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

Examples

			4484 is a term because 4484 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.
		

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, 9):
        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])

A345850 Numbers that are the sum of nine fourth powers in exactly eight ways.

Original entry on oeis.org

6804, 6869, 8019, 8084, 8324, 8499, 8564, 9044, 9124, 9219, 9234, 9284, 9364, 9429, 9474, 9494, 9604, 9669, 9749, 9779, 10148, 10259, 10293, 10339, 10388, 10453, 10514, 10579, 10628, 10644, 10754, 10789, 11029, 11059, 11189, 11204, 11299, 11363, 11364, 11379
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345592 at term 5 because 8259 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 8^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 9^4 = 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4.

Examples

			6869 is a term because 6869 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 9^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.
		

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, 9):
        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])

A345799 Numbers that are the sum of nine cubes in exactly seven ways.

Original entry on oeis.org

624, 629, 631, 650, 657, 687, 694, 707, 713, 720, 727, 746, 753, 755, 763, 768, 777, 779, 781, 784, 786, 789, 792, 796, 798, 803, 807, 820, 822, 824, 831, 833, 848, 849, 854, 870, 873, 875, 876, 879, 884, 885, 889, 890, 892, 898, 899, 901, 902, 904, 905, 906
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345546 at term 12 because 744 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 7^3 = 1^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 8^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3.
Likely finite.

Examples

			629 is a term because 629 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3.
		

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, 9):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v == 7])
        for x in range(len(rets)):
            print(rets[x])

A345859 Numbers that are the sum of ten fourth powers in exactly seven ways.

Original entry on oeis.org

4485, 5445, 5460, 5525, 5540, 5590, 5605, 5670, 5700, 5715, 5765, 5780, 5830, 5845, 6645, 6710, 6775, 6855, 6900, 6915, 6930, 6935, 6965, 6980, 7175, 7190, 7235, 7255, 7335, 7364, 7415, 7430, 7475, 7479, 7495, 7510, 7604, 7620, 7654, 7669, 7670, 7685, 7715
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345600 at term 16 because 6675 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4 = 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4.

Examples

			5445 is a term because 5445 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 = 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 8^4 = 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4.
		

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, 10):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v == 7])
        for x in range(len(rets)):
            print(rets[x])

A346342 Numbers that are the sum of nine fifth powers in exactly seven ways.

Original entry on oeis.org

1431429, 1439173, 1447570, 1504636, 1597929, 1671167, 1696159, 1697686, 1697928, 1778835, 1936454, 1975049, 2017344, 2092122, 2182161, 2198967, 2208680, 2280818, 2283911, 2289343, 2314335, 2329845, 2340319, 2345806, 2362370, 2388651, 2497771, 2529407, 2530672
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345624 at term 1 because 1431398 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 12^5 + 16^5 = 1^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.

Examples

			1431398 is a term because 1431398 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 12^5 + 16^5 = 1^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 1^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.
		

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, 9):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v == 7])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-7 of 7 results.