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.

A345539 Numbers that are the sum of eight cubes in nine or more ways.

Original entry on oeis.org

984, 1080, 1136, 1171, 1185, 1192, 1197, 1204, 1223, 1243, 1262, 1269, 1273, 1280, 1288, 1295, 1299, 1306, 1318, 1325, 1332, 1333, 1337, 1344, 1356, 1360, 1369, 1370, 1374, 1377, 1379, 1386, 1393, 1397, 1400, 1404, 1406, 1412, 1415, 1416, 1419, 1422, 1423
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1080 is a term because 1080 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3.
		

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, 8):
        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])

A345575 Numbers that are the sum of seven fourth powers in nine or more ways.

Original entry on oeis.org

19491, 21267, 21332, 23652, 31251, 35427, 36052, 37812, 38067, 39891, 40356, 41732, 41747, 43267, 43876, 43891, 43956, 44131, 44196, 44532, 44547, 44612, 45156, 45171, 45411, 45651, 45652, 45827, 45891, 45892, 45907, 46276, 46451, 46516, 47427, 47667, 47971
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			21267 is a term because 21267 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 2^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 7^4 + 7^4 + 8^4 + 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, 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])

A345583 Numbers that are the sum of eight fourth powers in eight or more ways.

Original entry on oeis.org

13268, 14212, 14788, 15427, 15667, 16612, 16627, 16692, 16707, 16772, 16822, 16852, 16882, 16947, 17348, 17363, 17428, 17493, 17877, 17972, 17987, 18052, 18117, 18227, 18948, 19157, 19237, 19252, 19267, 19412, 19492, 19507, 19572, 19682, 19747, 19748, 19828
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			14212 is a term because 14212 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 8^4 + 10^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 10^4 = 1^4 + 1^4 + 1^4 + 5^4 + 6^4 + 8^4 + 8^4 + 8^4 = 1^4 + 2^4 + 4^4 + 4^4 + 5^4 + 7^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 8^4 + 9^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 10^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 10^4 = 3^4 + 4^4 + 4^4 + 5^4 + 7^4 + 7^4 + 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, 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])

A345593 Numbers that are the sum of nine fourth powers in nine or more ways.

Original entry on oeis.org

8259, 9299, 9539, 10709, 10819, 10884, 10949, 10964, 11124, 11444, 11573, 11668, 11684, 11924, 12099, 12164, 12339, 12404, 12549, 12708, 12773, 12853, 12918, 12948, 13013, 13139, 13204, 13269, 13284, 13349, 13379, 13444, 13509, 13524, 13589, 13764, 13829
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			9299 is a term because 9299 = 1^4 + 1^4 + 1^4 + 2^4 + 6^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 7^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 9^4 = 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 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 >= 9])
        for x in range(len(rets)):
            print(rets[x])

A345841 Numbers that are the sum of eight fourth powers in exactly nine ways.

Original entry on oeis.org

15427, 16692, 17348, 17493, 18052, 18227, 19267, 19412, 19572, 19748, 20852, 21443, 21493, 21637, 21652, 21653, 21827, 21877, 21972, 22037, 22212, 22388, 22501, 22548, 22868, 22932, 23107, 23412, 23413, 23428, 23828, 23893, 23972, 24037, 24131, 24212, 24517
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345584 at term 5 because 17972 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 + 10^4 = 1^4 + 1^4 + 5^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 5^4 + 7^4 + 11^4 = 1^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 6^4 + 11^4 = 1^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 + 10^4 = 1^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 + 10^4 = 1^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 8^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 7^4 + 9^4 + 9^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 9^4 + 9^4.

Examples

			16692 is a term because 16692 = 1^4 + 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 + 10^4 = 1^4 + 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 11^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 10^4 = 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 = 1^4 + 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 5^4 + 7^4 + 9^4 + 9^4 = 2^4 + 3^4 + 4^4 + 5^4 + 6^4 + 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, 8):
        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])

A345585 Numbers that are the sum of eight fourth powers in ten or more ways.

Original entry on oeis.org

17972, 17987, 19492, 19507, 19747, 20116, 20787, 21268, 21283, 21333, 21348, 21413, 21508, 21523, 21588, 21892, 21957, 22067, 22132, 22563, 22628, 23172, 23237, 23252, 23587, 23588, 23603, 23653, 23668, 23733, 23843, 23908, 24277, 24452, 24802, 24948, 25363
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			17987 is a term because 17987 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 6^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 5^4 + 7^4 + 11^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 6^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 + 10^4 = 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 + 10^4 = 2^4 + 4^4 + 5^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 = 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 + 10^4 = 3^4 + 5^4 + 6^4 + 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, 8):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 10])
        for x in range(len(rets)):
            print(rets[x])

A345617 Numbers that are the sum of eight fifth powers in nine or more ways.

Original entry on oeis.org

8742208, 15539667, 18913169, 19987308, 20135313, 21505583, 21512966, 21563089, 21727552, 22237510, 22256608, 22438990, 22545600, 22686818, 22932525, 23106589, 23122550, 23189782, 23221517, 23287858, 23346048, 23477344, 23798742, 23847285, 23931325, 24138358
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			15539667 is a term because 15539667 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 24^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 24^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 22^5 + 24^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 19^5 + 20^5 + 24^5 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5.
		

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 >= 9])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-7 of 7 results.