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.

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])

A345616 Numbers that are the sum of eight fifth powers in eight or more ways.

Original entry on oeis.org

8625619, 8742208, 9773236, 10036233, 10071050, 12247994, 13180706, 13377868, 13662501, 14584992, 14591744, 14611077, 15251119, 15539667, 16112362, 16374250, 16391025, 16472544, 16588000, 16667851, 17059075, 17216298, 17405300, 17917097, 18107564, 18392902
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			8742208 is a term because 8742208 = 1^5 + 1^5 + 2^5 + 3^5 + 5^5 + 7^5 + 15^5 + 24^5 = 1^5 + 1^5 + 9^5 + 9^5 + 11^5 + 17^5 + 18^5 + 22^5 = 1^5 + 3^5 + 7^5 + 12^5 + 12^5 + 13^5 + 17^5 + 23^5 = 2^5 + 5^5 + 6^5 + 7^5 + 15^5 + 15^5 + 15^5 + 23^5 = 3^5 + 3^5 + 7^5 + 9^5 + 12^5 + 12^5 + 21^5 + 21^5 = 4^5 + 4^5 + 4^5 + 11^5 + 11^5 + 12^5 + 21^5 + 21^5 = 4^5 + 4^5 + 8^5 + 8^5 + 9^5 + 15^5 + 17^5 + 23^5 = 8^5 + 13^5 + 14^5 + 14^5 + 14^5 + 16^5 + 19^5 + 20^5 = 10^5 + 12^5 + 12^5 + 13^5 + 16^5 + 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, 8):
        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])

A345629 Numbers that are the sum of seven fifth powers in seven or more ways.

Original entry on oeis.org

28608832, 35663099, 36090526, 36620574, 46998599, 51095638, 52541851, 54233651, 54827543, 54886349, 61263643, 61634374, 63514593, 64810976, 65198607, 66708676, 67887843, 70979107, 72970305, 74002457, 74115801, 74132607, 74487093, 75044651, 75378359
Offset: 1

Views

Author

David Consiglio, Jr., Jun 22 2021

Keywords

Examples

			35663099 is a term because 35663099 = 1^5 + 9^5 + 16^5 + 17^5 + 24^5 + 24^5 + 28^5 = 2^5 + 3^5 + 17^5 + 23^5 + 24^5 + 24^5 + 26^5 = 2^5 + 10^5 + 15^5 + 17^5 + 23^5 + 23^5 + 29^5 = 4^5 + 8^5 + 13^5 + 19^5 + 21^5 + 27^5 + 27^5 = 4^5 + 11^5 + 13^5 + 19^5 + 20^5 + 22^5 + 30^5 = 5^5 + 6^5 + 19^5 + 19^5 + 20^5 + 20^5 + 30^5 = 7^5 + 9^5 + 12^5 + 18^5 + 18^5 + 27^5 + 28^5.
		

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 >= 7])
        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])

A346285 Numbers that are the sum of seven fifth powers in exactly eight ways.

Original entry on oeis.org

36620574, 80552143, 81401376, 82078424, 92347417, 93653176, 94486699, 94626949, 98873875, 105674625, 121050874, 125959393, 129228307, 144209018, 145340799, 147245218, 147898763, 151727082, 151923168, 152361276, 152664876, 153877208, 155107349, 155270357
Offset: 1

Views

Author

David Consiglio, Jr., Jul 12 2021

Keywords

Comments

Differs from A345630 at term 11 because 110276376 = 1^5 + 3^5 + 5^5 + 7^5 + 17^5 + 23^5 + 40^5 = 5^5 + 10^5 + 16^5 + 16^5 + 19^5 + 20^5 + 40^5 = 1^5 + 8^5 + 14^5 + 16^5 + 21^5 + 27^5 + 39^5 = 7^5 + 8^5 + 11^5 + 14^5 + 16^5 + 33^5 + 37^5 = 4^5 + 7^5 + 8^5 + 13^5 + 26^5 + 31^5 + 37^5 = 1^5 + 5^5 + 6^5 + 20^5 + 28^5 + 29^5 + 37^5 = 3^5 + 3^5 + 7^5 + 18^5 + 27^5 + 32^5 + 36^5 = 6^5 + 12^5 + 18^5 + 25^5 + 30^5 + 31^5 + 34^5 = 6^5 + 10^5 + 20^5 + 27^5 + 27^5 + 33^5 + 33^5.

Examples

			36620574 is a term because 36620574 = 4^5 + 9^5 + 14^5 + 17^5 + 18^5 + 21^5 + 31^5 = 1^5 + 12^5 + 13^5 + 14^5 + 20^5 + 24^5 + 30^5 = 8^5 + 9^5 + 12^5 + 13^5 + 16^5 + 27^5 + 29^5 = 5^5 + 7^5 + 7^5 + 20^5 + 23^5 + 23^5 + 29^5 = 17^5 + 18^5 + 20^5 + 20^5 + 20^5 + 20^5 + 29^5 = 2^5 + 7^5 + 14^5 + 14^5 + 23^5 + 26^5 + 28^5 = 4^5 + 8^5 + 8^5 + 17^5 + 23^5 + 27^5 + 27^5 = 2^5 + 3^5 + 14^5 + 18^5 + 24^5 + 26^5 + 27^5.
		

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

A345722 Numbers that are the sum of six fifth powers in eight or more ways.

Original entry on oeis.org

2295937600, 4335900525, 6251954544, 8986552608, 9085584992, 13413708308, 14539246326, 15277569450, 15728636000, 16770321920, 16873011232, 16933805856, 17572402769, 17713454592, 17960776999, 18190647200, 19621666592, 20570070125, 20827689300
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			4335900525 is a term because 4335900525 = 2^5 + 24^5 + 34^5 + 56^5 + 61^5 + 78^5 = 3^5 + 21^5 + 37^5 + 54^5 + 62^5 + 78^5 = 3^5 + 21^5 + 39^5 + 49^5 + 66^5 + 77^5 = 3^5 + 26^5 + 32^5 + 49^5 + 72^5 + 73^5 = 8^5 + 16^5 + 42^5 + 49^5 + 61^5 + 79^5 = 9^5 + 13^5 + 43^5 + 47^5 + 66^5 + 77^5 = 19^5 + 20^5 + 30^5 + 45^5 + 61^5 + 80^5 = 21^5 + 24^5 + 28^5 + 37^5 + 67^5 + 78^5.
		

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 >= 8])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.