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.

A345563 Numbers that are the sum of six fourth powers in six or more ways.

Original entry on oeis.org

21251, 37811, 38051, 43251, 43571, 43875, 44115, 44531, 45155, 45651, 45891, 47411, 47586, 48276, 49796, 49971, 52195, 53235, 53315, 54131, 56290, 57395, 57460, 57570, 58035, 58500, 59075, 59330, 59780, 59795, 59811, 59860, 60035, 62180, 62211, 63971, 66340
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			37811 is a term because 37811 = 1^4 + 2^4 + 2^4 + 7^4 + 11^4 + 12^4 = 2^4 + 2^4 + 4^4 + 7^4 + 9^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 9^4 + 13^4 = 3^4 + 4^4 + 8^4 + 8^4 + 11^4 + 11^4 = 4^4 + 6^4 + 7^4 + 9^4 + 9^4 + 12^4 = 5^4 + 5^4 + 9^4 + 10^4 + 10^4 + 10^4.
		

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 >= 6])
        for x in range(len(rets)):
            print(rets[x])

A345817 Numbers that are the sum of six fourth powers in exactly five ways.

Original entry on oeis.org

15395, 16610, 18866, 19235, 19410, 20996, 21011, 21316, 21331, 21491, 21620, 23811, 25091, 29700, 29715, 29906, 29955, 30356, 30995, 31235, 31266, 31331, 31506, 32035, 33651, 33795, 33891, 35171, 35411, 35636, 35796, 35971, 37971, 38595, 38675, 39266, 39890
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345562 at term 8 because 21251 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4.

Examples

			16610 is a term because 16610 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 5^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 10^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 5^4 + 6^4 + 7^4 + 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, 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])

A345819 Numbers that are the sum of six fourth powers in exactly seven ways.

Original entry on oeis.org

21251, 43875, 48276, 49796, 53315, 58500, 59795, 59811, 67875, 68306, 69155, 69779, 71955, 72051, 72131, 73970, 74420, 74851, 77010, 80291, 80515, 81875, 82275, 84515, 86436, 86451, 86531, 87075, 88355, 88660, 88675, 90355, 91475, 93410, 93650, 94690, 95155
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345564 at term 6 because 58035 = 1^4 + 1^4 + 9^4 + 10^4 + 12^4 + 12^4 = 1^4 + 4^4 + 5^4 + 8^4 + 11^4 + 14^4 = 1^4 + 5^4 + 6^4 + 11^4 + 12^4 + 12^4 = 2^4 + 2^4 + 4^4 + 5^4 + 13^4 + 13^4 = 2^4 + 6^4 + 6^4 + 7^4 + 7^4 + 15^4 = 2^4 + 8^4 + 10^4 + 11^4 + 11^4 + 11^4 = 3^4 + 4^4 + 4^4 + 4^4 + 9^4 + 15^4 = 4^4 + 5^4 + 6^4 + 9^4 + 12^4 + 13^4.

Examples

			43875 is a term because 43875 = 1^4 + 2^4 + 9^4 + 9^4 + 10^4 + 12^4 = 2^4 + 2^4 + 2^4 + 5^4 + 11^4 + 13^4 = 2^4 + 2^4 + 5^4 + 7^4 + 7^4 + 14^4 = 2^4 + 5^4 + 6^4 + 9^4 + 11^4 + 12^4 = 3^4 + 7^4 + 8^4 + 9^4 + 10^4 + 12^4 = 4^4 + 4^4 + 7^4 + 7^4 + 10^4 + 13^4 = 5^4 + 7^4 + 8^4 + 8^4 + 8^4 + 13^4.
		

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

A345828 Numbers that are the sum of seven fourth powers in exactly six ways.

Original entry on oeis.org

10787, 15396, 15411, 15586, 15651, 16611, 16626, 16676, 16866, 17956, 18867, 19156, 19236, 19251, 19411, 19426, 19666, 20035, 20771, 21012, 21187, 21397, 21412, 21442, 21492, 21572, 21621, 21811, 21891, 22116, 22132, 22292, 22307, 22372, 22595, 22660, 22962
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345572 at term 9 because 16691 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4.

Examples

			15396 is a term because 15396 = 1^4 + 1^4 + 1^4 + 1^4 + 6^4 + 8^4 + 10^4 = 1^4 + 1^4 + 2^4 + 5^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 11^4 = 1^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 10^4 = 1^4 + 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 3^4 + 4^4 + 5^4 + 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, 7):
        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])

A344941 Numbers that are the sum of five fourth powers in exactly six ways.

Original entry on oeis.org

151300, 225890, 236194, 243235, 246674, 250834, 286114, 288579, 300835, 302130, 302210, 303235, 309059, 317795, 320195, 334819, 334899, 335443, 336210, 338914, 346835, 356899, 363379, 366995, 373234, 375619, 389875, 391154, 392259, 393314, 394354, 412339
Offset: 1

Views

Author

David Consiglio, Jr., Jun 03 2021

Keywords

Comments

Differs from A344940 at term 2 because 197779 = 1^4 + 5^4 + 6^4 + 16^4 + 19^4 = 1^4 + 7^4 + 11^4 + 12^4 + 20^4 = 1^4 + 10^4 + 12^4 + 17^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 21^4 = 3^4 + 5^4 + 6^4 + 6^4 + 21^4 = 4^4 + 7^4 + 9^4 + 13^4 + 20^4 = 11^4 + 13^4 + 14^4 + 15^4 + 16^4.

Examples

			151300 is a term because 151300 = 3^4 + 3^4 + 3^4 + 12^4 + 19^4  = 3^4 + 11^4 + 11^4 + 14^4 + 17^4  = 3^4 + 13^4 + 13^4 + 13^4 + 16^4  = 6^4 + 9^4 + 9^4 + 9^4 + 19^4  = 7^4 + 11^4 + 11^4 + 11^4 + 18^4  = 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, 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])

A345768 Numbers that are the sum of six cubes in exactly six ways.

Original entry on oeis.org

1377, 1488, 1586, 1595, 1647, 1673, 1677, 1738, 1764, 1799, 1829, 1836, 1837, 1862, 1881, 1890, 1911, 1953, 1955, 2007, 2011, 2014, 2018, 2025, 2044, 2070, 2079, 2097, 2107, 2108, 2142, 2153, 2170, 2177, 2203, 2214, 2216, 2222, 2223, 2226, 2229, 2252, 2258
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345515 at term 8 because 1710 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 9^3 = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 9^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 10^3 = 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 10^3 = 2^3 + 3^3 + 4^3 + 4^3 + 6^3 + 11^3 = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.

Examples

			1488 is a term because 1488 = 1^3 + 1^3 + 1^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.
		

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 == 6])
        for x in range(len(rets)):
            print(rets[x])

A346361 Numbers that are the sum of six fifth powers in exactly six ways.

Original entry on oeis.org

287718651, 553545456, 746783675, 972232800, 1005620508, 1040741042, 1070652352, 1074892544, 1182426366, 1197332400, 1243267146, 1317183650, 1364866263, 1387455091, 1429663400, 1498160992, 1529189818, 1554833117, 1558594400, 1610298901, 1623782765, 1627228231
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345720 at term 10 because 1184966816 = 15^5 + 24^5 + 27^5 + 38^5 + 39^5 + 63^5 = 2^5 + 28^5 + 36^5 + 36^5 + 42^5 + 62^5 = 4^5 + 24^5 + 38^5 + 38^5 + 40^5 + 62^5 = 21^5 + 32^5 + 37^5 + 41^5 + 45^5 + 60^5 = 8^5 + 14^5 + 34^5 + 40^5 + 52^5 + 58^5 = 11^5 + 17^5 + 22^5 + 49^5 + 51^5 + 56^5 = 11^5 + 16^5 + 22^5 + 52^5 + 52^5 + 53^5.

Examples

			287718651 is a term because 287718651 = 10^5 + 11^5 + 20^5 + 22^5 + 30^5 + 48^5 = 8^5 + 10^5 + 21^5 + 27^5 + 27^5 + 48^5 = 3^5 + 6^5 + 25^5 + 30^5 + 30^5 + 47^5 = 9^5 + 10^5 + 13^5 + 26^5 + 37^5 + 46^5 = 6^5 + 9^5 + 14^5 + 31^5 + 35^5 + 46^5 = 10^5 + 11^5 + 12^5 + 23^5 + 41^5 + 44^5.
		

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 == 6])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-7 of 7 results.