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.

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])

A345540 Numbers that are the sum of eight cubes in ten or more ways.

Original entry on oeis.org

1185, 1243, 1262, 1288, 1295, 1299, 1386, 1393, 1397, 1400, 1412, 1419, 1423, 1448, 1449, 1451, 1458, 1460, 1464, 1467, 1475, 1477, 1497, 1501, 1503, 1504, 1505, 1512, 1513, 1514, 1516, 1521, 1523, 1539, 1540, 1541, 1542, 1553, 1558, 1559, 1560, 1565, 1566
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1243 is a term because 1243 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 8^3 = 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 7^3.

Links

Crossrefs

Cf. A345497, A345506, A345539, A345549, A345585, A345792.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])

A345576 Numbers that are the sum of seven fourth powers in ten or more ways.

Original entry on oeis.org

31251, 44547, 45827, 45892, 45907, 47667, 47971, 48292, 49572, 49812, 50052, 51092, 52372, 53316, 53476, 54531, 54596, 54756, 54996, 57411, 58036, 58116, 58276, 58516, 58660, 58756, 59331, 59781, 59796, 59811, 59827, 59861, 59876, 59892, 60036, 60371, 60436
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			44547 is a term because 44547 = 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 11^4 + 13^4 = 1^4 + 2^4 + 2^4 + 6^4 + 7^4 + 7^4 + 14^4 = 1^4 + 2^4 + 6^4 + 6^4 + 9^4 + 11^4 + 12^4 = 1^4 + 6^4 + 7^4 + 8^4 + 8^4 + 8^4 + 13^4 = 2^4 + 2^4 + 8^4 + 9^4 + 9^4 + 9^4 + 12^4 = 2^4 + 4^4 + 6^4 + 6^4 + 9^4 + 9^4 + 13^4 = 2^4 + 4^4 + 7^4 + 7^4 + 8^4 + 11^4 + 12^4 = 3^4 + 3^4 + 4^4 + 4^4 + 7^4 + 12^4 + 12^4 = 3^4 + 6^4 + 6^4 + 7^4 + 8^4 + 11^4 + 12^4 = 4^4 + 4^4 + 8^4 + 8^4 + 9^4 + 11^4 + 11^4.

Links

Crossrefs

Cf. A345506, A345567, A345575, A345585, A345643, A345832.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])

A345594 Numbers that are the sum of nine fourth powers in ten or more ways.

Original entry on oeis.org

9299, 12708, 12948, 13269, 13349, 13524, 13589, 13764, 13829, 13893, 14133, 14228, 14468, 14564, 14804, 14869, 14934, 14964, 15014, 15044, 15094, 15109, 15174, 15189, 15333, 15413, 15428, 15429, 15443, 15508, 15524, 15573, 15588, 15604, 15653, 15669, 15683
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			12708 is a term because 12708 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 10^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 10^4 = 1^4 + 1^4 + 1^4 + 5^4 + 6^4 + 6^4 + 6^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 6^4 + 8^4 + 9^4 = 1^4 + 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 6^4 + 7^4 + 9^4 = 1^4 + 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 6^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 10^4 = 2^4 + 2^4 + 3^4 + 3^4 + 5^4 + 6^4 + 7^4 + 8^4 + 8^4 = 2^4 + 4^4 + 4^4 + 4^4 + 5^4 + 7^4 + 7^4 + 7^4 + 8^4 = 3^4 + 4^4 + 4^4 + 5^4 + 6^4 + 6^4 + 7^4 + 7^4 + 8^4.

Links

Crossrefs

Cf. A345549, A345585, A345593, A345603, A345627, A345852.

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 >= 10])
    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])

A345618 Numbers that are the sum of eight fifth powers in ten or more ways.

Original entry on oeis.org

15539667, 22932525, 24393600, 24650406, 24952961, 24953742, 25054306, 25142513, 25201550, 25423794, 26001294, 26851552, 27396567, 27988486, 28609075, 29309819, 29558650, 31052406, 31794336, 32223105, 32527286, 32610600, 32807777, 32890541, 32998317, 33015125
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

Cf. A345585, A345617, A345627, A345643, A346335.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-6 of 6 results.

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