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 1-6 of 6 results.

A345562 Numbers that are the sum of six fourth powers in five or more ways.

Original entry on oeis.org

15395, 16610, 18866, 19235, 19410, 20996, 21011, 21251, 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, 37811, 37971, 38051, 38595
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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])

A345863 Numbers that are the sum of five fifth powers in five or more ways.

Original entry on oeis.org

9006349824, 65799210368, 67629776576, 181085909632, 188189635424, 288203194368, 295677350451, 467139768468, 471359089024, 656243139157, 691381929281, 797466940832, 854533526901, 874953049024, 891862586132, 953769598750, 1038549256768
Offset: 1

Views

Author

David Consiglio, Jr., Jun 27 2021

Keywords

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])

A345608 Numbers that are the sum of seven fifth powers in five or more ways.

Original entry on oeis.org

6768576, 6776120, 7883668, 8625376, 8740709, 10036201, 10604054, 12476826, 12618493, 13006575, 13060213, 13080706, 13174250, 13536416, 13550162, 13562501, 13662500, 14110656, 14583968, 15169276, 15247994, 16053313, 16060683, 16374218, 16573507, 16600001
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			6776120 is a term because 6776120 = 2^5 + 4^5 + 7^5 + 12^5 + 17^5 + 18^5 + 20^5 = 3^5 + 6^5 + 6^5 + 12^5 + 14^5 + 18^5 + 21^5 = 4^5 + 6^5 + 8^5 + 11^5 + 13^5 + 16^5 + 22^5 = 4^5 + 7^5 + 7^5 + 7^5 + 16^5 + 19^5 + 20^5 = 5^5 + 6^5 + 6^5 + 8^5 + 16^5 + 19^5 + 20^5.
		

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 >= 5])
        for x in range(len(rets)):
            print(rets[x])

A345718 Numbers that are the sum of six fifth powers in four or more ways.

Original entry on oeis.org

12047994, 20646208, 21017489, 21300963, 21741819, 24993485, 27669050, 28576064, 30193856, 30785920, 35480456, 35735194, 36082750, 37303264, 39035975, 46814942, 47963291, 50047062, 50724345, 52987561, 53076800, 53606848, 54827300, 55101101, 56766906
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			20646208 is a term because 20646208 = 2^5 + 12^5 + 12^5 + 16^5 + 18^5 + 28^5 = 3^5 + 4^5 + 4^5 + 8^5 + 10^5 + 29^5 = 6^5 + 6^5 + 12^5 + 14^5 + 24^5 + 26^5 = 7^5 + 7^5 + 8^5 + 16^5 + 25^5 + 25^5.
		

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 >= 4])
        for x in range(len(rets)):
            print(rets[x])

A345720 Numbers that are the sum of six fifth powers in six or more ways.

Original entry on oeis.org

287718651, 553545456, 746783675, 972232800, 1005620508, 1040741042, 1070652352, 1074892544, 1182426366, 1184966816, 1197332400, 1243267146, 1317183650, 1364866263, 1387455091, 1429663400, 1498160992, 1529189818, 1554833117, 1558594400, 1610298901
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			553545456 is a term because 553545456 = 1^5 + 14^5 + 20^5 + 24^5 + 47^5 + 50^5 = 4^5 + 14^5 + 37^5 + 42^5 + 43^5 + 46^5 = 4^5 + 26^5 + 29^5 + 34^5 + 42^5 + 51^5 = 9^5 + 15^5 + 22^5 + 22^5 + 33^5 + 55^5 = 9^5 + 26^5 + 29^5 + 32^5 + 37^5 + 53^5 = 12^5 + 24^5 + 27^5 + 32^5 + 38^5 + 53^5.
		

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])

A346360 Numbers that are the sum of six fifth powers in exactly five ways.

Original entry on oeis.org

54827300, 74115800, 74883600, 75609125, 113088250, 120274275, 166078869, 169692136, 174781858, 178736448, 182341225, 185558208, 194939538, 203054589, 218814275, 235067008, 250989825, 251772882, 252721458, 255453233, 258124975, 274616694, 282859667, 287677700
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345719 at term 25 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.

Examples

			54827300 is a term because 54827300 = 4^5 + 7^5 + 21^5 + 22^5 + 23^5 + 33^5 = 5^5 + 10^5 + 15^5 + 20^5 + 28^5 + 32^5 = 1^5 + 14^5 + 16^5 + 19^5 + 28^5 + 32^5 = 4^5 + 11^5 + 13^5 + 22^5 + 29^5 + 31^5 = 5^5 + 6^5 + 19^5 + 20^5 + 29^5 + 31^5.
		

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 == 5])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.