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.

A345584 Numbers that are the sum of eight fourth powers in nine or more ways.

Original entry on oeis.org

15427, 16692, 17348, 17493, 17972, 17987, 18052, 18227, 19267, 19412, 19492, 19507, 19572, 19747, 19748, 20116, 20787, 20852, 21268, 21283, 21333, 21348, 21413, 21443, 21493, 21508, 21523, 21588, 21637, 21652, 21653, 21827, 21877, 21892, 21957, 21972, 22037
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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.

Links

Crossrefs

Cf. A345539, A345575, A345583, A345585, A345593, A345617, A345841.

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])

A345831 Numbers that are the sum of seven fourth powers in exactly nine ways.

Original entry on oeis.org

19491, 21267, 21332, 23652, 35427, 36052, 37812, 38067, 39891, 40356, 41732, 41747, 43267, 43876, 43891, 43956, 44131, 44196, 44532, 44612, 45156, 45171, 45411, 45651, 45652, 45891, 46276, 46451, 46516, 47427, 48036, 48052, 48532, 48707, 49747, 49956, 49987
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345575 at term 5 because 31251 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 10^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 10^4 + 11^4 = 1^4 + 4^4 + 4^4 + 4^4 + 5^4 + 6^4 + 13^4 = 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 5^4 + 6^4 + 11^4 + 11^4 = 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 10^4 + 11^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 10^4 + 12^4 = 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 10^4 = 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 10^4 + 11^4 = 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 11^4.

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.

Links

Crossrefs

Cf. A345575, A345781, A345821, A345830, A345832, A345841, A346286.

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])

A345840 Numbers that are the sum of eight fourth powers in exactly eight ways.

Original entry on oeis.org

13268, 14212, 14788, 15667, 16612, 16627, 16707, 16772, 16822, 16852, 16882, 16947, 17363, 17428, 17877, 18117, 18948, 19157, 19237, 19252, 19682, 19828, 20291, 20372, 20612, 20707, 20722, 20772, 20917, 20962, 21253, 21331, 21458, 21478, 21573, 21717, 21763
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345583 at term 4 because 15427 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 6^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 5^4 + 8^4 + 8^4 + 9^4 = 1^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 11^4 = 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 10^4 = 2^4 + 2^4 + 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 = 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 + 10^4 = 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 10^4 = 4^4 + 4^4 + 5^4 + 6^4 + 7^4 + 7^4 + 8^4 + 8^4.

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.

Links

Crossrefs

Cf. A345583, A345790, A345830, A345839, A345841, A345850, A346333.

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])

A345851 Numbers that are the sum of nine fourth powers in exactly nine ways.

Original entry on oeis.org

8259, 9539, 10709, 10819, 10884, 10949, 10964, 11124, 11444, 11573, 11668, 11684, 11924, 12099, 12164, 12339, 12404, 12549, 12773, 12853, 12918, 13013, 13139, 13204, 13284, 13379, 13444, 13509, 13958, 13988, 14053, 14213, 14293, 14308, 14373, 14403, 14484
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345593 at term 2 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.

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.

Links

Crossrefs

Cf. A345593, A345801, A345841, A345850, A345852, A345861, A346344.

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])

A345791 Numbers that are the sum of eight cubes in exactly nine ways.

Original entry on oeis.org

984, 1080, 1136, 1171, 1192, 1197, 1204, 1223, 1269, 1273, 1280, 1306, 1318, 1325, 1332, 1333, 1337, 1344, 1356, 1360, 1369, 1370, 1374, 1377, 1379, 1404, 1406, 1415, 1416, 1422, 1425, 1430, 1432, 1438, 1442, 1444, 1445, 1456, 1476, 1481, 1486, 1488, 1494
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345539 at term 5 because 1185 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 6^3 + 8^3 = 1^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 + 9^3 = 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 9^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 8^3 + 8^3 = 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 + 7^3 + 7^3.
Likely finite.

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.

Links

Crossrefs

Cf. A345539, A345781, A345790, A345792, A345801, A345841.

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])

A345842 Numbers that are the sum of eight fourth powers in exactly ten ways.

Original entry on oeis.org

17972, 17987, 19492, 19507, 19747, 20116, 21283, 21333, 21413, 21508, 21588, 22067, 22563, 23237, 23252, 23587, 23588, 23603, 23653, 24277, 24452, 24802, 24948, 25603, 26228, 27347, 27683, 27813, 27893, 27973, 28532, 28852, 28853, 28933, 29108, 29173, 29491
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345585 at term 7 because 20787 = 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 8^4 + 8^4 + 10^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 + 9^4 + 10^4 = 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 6^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 + 8^4 + 10^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4 + 11^4 = 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

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.

Links

Crossrefs

Cf. A345585, A345792, A345832, A345841, A345852, A346335.

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])

A346334 Numbers that are the sum of eight fifth powers in exactly nine ways.

Original entry on oeis.org

8742208, 18913169, 19987308, 20135313, 21505583, 21512966, 21563089, 21727552, 22237510, 22256608, 22438990, 22545600, 22686818, 23106589, 23122550, 23189782, 23221517, 23287858, 23346048, 23477344, 23798742, 23847285, 23931325, 24138358, 24385108, 24394139
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345617 at term 2 because 15539667 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 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 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 24^5 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5.

Examples

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

Links

Crossrefs

Cf. A345617, A345841, A346286, A346333, A346335, A346344.

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.

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