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.

A344355 Numbers that are the sum of five fourth powers in exactly four ways.

Original entry on oeis.org

20995, 21235, 31250, 41474, 43235, 43250, 43315, 43490, 43859, 45139, 46290, 47570, 51939, 53234, 53299, 54994, 56274, 57379, 57410, 57779, 59329, 63970, 67010, 68035, 68290, 71795, 71954, 73730, 73954, 75714, 75794, 77890, 82099, 84499, 86275, 86450, 87730, 92500, 93474, 93859, 94130, 94210, 96194
Offset: 1

Views

Author

David Consiglio, Jr., May 15 2021

Keywords

Comments

Differs from A344354 at term 22 because 59779 = 1^4 + 1^4 + 5^4 + 12^4 + 14^4 = 1^4 + 6^4 + 6^4 + 9^4 + 15^4 = 2^4 + 9^4 + 10^4 + 11^4 + 13^4 = 4^4 + 7^4 + 7^4 + 8^4 + 15^4 = 7^4 + 7^4 + 9^4 + 10^4 + 14^4.

Examples

			31250 is a term of this sequence because 31250 = 2^4 + 2^4 + 4^4 + 7^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 13^4 = 4^4 + 6^4 + 7^4 + 9^4 + 12^4 = 5^4 + 5^4 + 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, 50)]
    for pos in cwr(power_terms, 5):
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 4])
    for x in range(len(rets)):
        print(rets[x])

A344358 Numbers that are the sum of five fourth powers in five or more ways.

Original entry on oeis.org

59779, 67859, 93394, 108274, 112850, 136915, 142354, 151300, 151475, 161459, 168979, 181219, 183539, 183604, 185299, 187699, 189394, 193379, 195394, 197779, 199090, 199474, 200979, 201874, 202979, 203299, 205859, 211059, 211330, 212419, 213730, 217154, 217810, 217890, 221779, 223090, 223155
Offset: 1

Views

Author

David Consiglio, Jr., May 15 2021

Keywords

Examples

			93394 is a term of this sequence because 93394 = 1^4 + 4^4 + 8^4 + 14^4 + 15^4 = 1^4 + 6^4 + 12^4 + 12^4 + 15^4 = 1^4 + 9^4 + 10^4 + 14^4 + 14^4 = 5^4 + 6^4 + 11^4 + 14^4 + 14^4 = 5^4 + 7^4 + 8^4 + 12^4 + 16^4.
		

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, 50)]
    for pos in cwr(power_terms, 5):
        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])

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])

A343988 Numbers that are the sum of five positive cubes in exactly five ways.

Original entry on oeis.org

1765, 1980, 2043, 2104, 2195, 2250, 2449, 2486, 2491, 2493, 2547, 2584, 2592, 2738, 2745, 2764, 2817, 2888, 2915, 2953, 2969, 3095, 3096, 3133, 3142, 3186, 3188, 3240, 3275, 3277, 3310, 3366, 3403, 3422, 3459, 3464, 3466, 3483, 3529, 3583, 3608, 3627, 3653, 3664, 3671, 3690, 3697, 3707, 3725, 3744, 3746, 3781
Offset: 1

Views

Author

David Consiglio, Jr., May 06 2021

Keywords

Comments

Differs from A343989 at term 7 because 2430 = 1^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 4^3 + 5^3 + 8^3 + 12^3 = 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 4^3 + 5^3 + 8^3 + 9^3 + 10^3.

Examples

			2043 is a term because 2043 = 1^3 + 4^3 + 5^3 + 5^3 + 12^3 = 2^3 + 2^3 + 3^3 + 10^3 + 10^3 = 2^3 + 3^3 + 4^3 + 6^3 + 12^3 = 4^3 + 5^3 + 5^3 + 9^3 + 10^3 = 4^3 + 6^3 + 6^3 + 6^3 + 11^3.
		

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(1,50)]
    for pos in cwr(power_terms,5):
        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])

A344357 Numbers that are the sum of four fourth powers in exactly five ways.

Original entry on oeis.org

2147874, 2266338, 2690658, 3189603, 3464178, 3754674, 4030419, 4165794, 4457298, 4884114, 5229378, 5978883, 5980178, 5981283, 6014178, 6134994, 6258723, 6313953, 6400194, 6612354, 7088898, 7498323, 7811874, 7918498, 8064018, 8292323, 8630259, 9146034, 9269523, 9388978, 9397683, 9726978
Offset: 1

Views

Author

David Consiglio, Jr., May 15 2021

Keywords

Comments

Differs from A344356 at term 7 because 3847554 = 2^4 + 13^4 + 29^4 + 42^4 = 2^4 + 21^4 + 22^4 + 43^4 = 6^4 + 11^4 + 17^4 + 44^4 = 6^4 + 31^4 + 32^4 + 37^4 = 9^4 + 29^4 + 32^4 + 38^4 = 13^4 + 26^4 + 32^4 + 39^4.

Examples

			2690658 is a term of this sequence because 2690658 = 2^4 + 8^4 + 33^4 + 35^4 = 3^4 + 4^4 + 19^4 + 40^4 = 7^4 + 8^4 + 30^4 + 37^4 = 9^4 + 21^4 + 30^4 + 36^4 = 16^4 + 25^4 + 32^4 + 33^4.
		

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, 50)]
    for pos in cwr(power_terms, 4):
        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])

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])

A346257 Numbers that are the sum of five fifth powers in exactly five ways.

Original entry on oeis.org

9006349824, 65799210368, 67629776576, 181085909632, 188189635424, 295677350451, 467139768468, 471359089024, 656243139157, 691381929281, 797466940832, 854533526901, 874953049024, 891862586132, 953769598750, 1038549256768, 1092458681568, 1182658308657
Offset: 1

Views

Author

David Consiglio, Jr., Jul 11 2021

Keywords

Comments

Differs from 103 terms known at term 6 because 288203194368 = 48^5 + 84^5 + 96^5 + 108^5 + 192^5 = 16^5 + 99^5 + 103^5 + 121^5 + 189^5 = 42^5 + 68^5 + 86^5 + 148^5 + 184^5 = 16^5 + 68^5 + 124^5 + 136^5 + 184^5 = 16^5 + 82^5 + 94^5 + 158^5 + 178^5 = 24^5 + 36^5 + 144^5 + 156^5 + 168^5.

Examples

			9006349824 is a term because 9006349824 = 24^5 + 42^5 + 48^5 + 54^5 + 96^5 = 21^5 + 34^5 + 43^5 + 74^5 + 92^5 = 8^5 + 34^5 + 62^5 + 68^5 + 92^5 = 8^5 + 41^5 + 47^5 + 79^5 + 89^5 = 12^5 + 18^5 + 72^5 + 78^5 + 84^5.
		

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, 5):
        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])
Showing 1-7 of 7 results.