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 1-5 of 5 results.

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])

A345626 Numbers that are the sum of nine fifth powers in nine or more ways.

Original entry on oeis.org

1969221, 2596936, 3353186, 3378178, 3923426, 3981447, 4094027, 4096729, 4112329, 4114188, 4129465, 4137209, 4147736, 4157156, 4170112, 4172994, 4254304, 4303773, 4410482, 4475846, 4477936, 4483379, 4485480, 4492410, 4501441, 4510461, 4543232, 4652011, 4691855
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			2596936 is a term because 2596936 = 1^5 + 1^5 + 4^5 + 5^5 + 9^5 + 13^5 + 13^5 + 13^5 + 17^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 17^5 = 1^5 + 4^5 + 7^5 + 7^5 + 7^5 + 9^5 + 9^5 + 14^5 + 18^5 = 1^5 + 5^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 12^5 + 13^5 + 14^5 + 17^5 = 2^5 + 5^5 + 5^5 + 6^5 + 6^5 + 6^5 + 15^5 + 15^5 + 16^5 = 3^5 + 3^5 + 5^5 + 6^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 11^5 + 11^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 8^5 + 9^5 + 16^5 + 17^5.

Links

Crossrefs

Cf. A345593, A345617, A345625, A345627, A345641, 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, 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])

A346345 Numbers that are the sum of nine fifth powers in exactly ten ways.

Original entry on oeis.org

4157156, 4492410, 4510461, 4915538, 5005474, 5015506, 5179747, 5219655, 5756794, 6323426, 6326519, 6382443, 6423394, 6705284, 6793170, 6861218, 7101038, 7147645, 7147656, 7148679, 7266240, 7280391, 7283268, 7314187, 7413493, 7422352, 7531076, 7651645, 7693425
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345627 at term 5 because 4948274 = 2^5 + 4^5 + 4^5 + 5^5 + 6^5 + 8^5 + 9^5 + 15^5 + 21^5 = 1^5 + 3^5 + 5^5 + 5^5 + 8^5 + 8^5 + 8^5 + 15^5 + 21^5 = 1^5 + 2^5 + 2^5 + 5^5 + 5^5 + 11^5 + 11^5 + 17^5 + 20^5 = 8^5 + 9^5 + 9^5 + 10^5 + 10^5 + 10^5 + 12^5 + 16^5 + 20^5 = 4^5 + 8^5 + 8^5 + 10^5 + 12^5 + 12^5 + 15^5 + 16^5 + 19^5 = 4^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 15^5 + 19^5 = 4^5 + 4^5 + 9^5 + 13^5 + 13^5 + 13^5 + 14^5 + 15^5 + 19^5 = 6^5 + 6^5 + 9^5 + 9^5 + 12^5 + 12^5 + 14^5 + 18^5 + 18^5 = 1^5 + 8^5 + 8^5 + 12^5 + 12^5 + 14^5 + 14^5 + 17^5 + 18^5 = 1^5 + 8^5 + 9^5 + 9^5 + 13^5 + 14^5 + 16^5 + 17^5 + 17^5 = 3^5 + 7^5 + 7^5 + 10^5 + 12^5 + 16^5 + 16^5 + 16^5 + 17^5.

Examples

			4157156 is a term because 4157156 = 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 9^5 + 21^5 = 1^5 + 1^5 + 3^5 + 4^5 + 5^5 + 5^5 + 8^5 + 8^5 + 21^5 = 1^5 + 4^5 + 4^5 + 8^5 + 10^5 + 12^5 + 12^5 + 16^5 + 19^5 = 1^5 + 4^5 + 4^5 + 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 19^5 = 5^5 + 5^5 + 5^5 + 5^5 + 7^5 + 9^5 + 15^5 + 17^5 + 18^5 = 3^5 + 3^5 + 5^5 + 6^5 + 9^5 + 10^5 + 16^5 + 16^5 + 18^5 = 1^5 + 1^5 + 5^5 + 5^5 + 13^5 + 13^5 + 15^5 + 15^5 + 18^5 = 2^5 + 3^5 + 4^5 + 4^5 + 10^5 + 14^5 + 16^5 + 16^5 + 17^5 = 11^5 + 11^5 + 12^5 + 12^5 + 12^5 + 12^5 + 13^5 + 16^5 + 17^5 = 2^5 + 2^5 + 2^5 + 5^5 + 12^5 + 15^5 + 16^5 + 16^5 + 16^5.

Links

Crossrefs

Cf. A345627, A345852, A346335, A346344, A346355.

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, 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])

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])

A345642 Numbers that are the sum of ten fifth powers in ten or more ways.

Original entry on oeis.org

1431641, 1439416, 1464377, 1464408, 1505660, 1531398, 1531640, 1564165, 1697929, 1703935, 1782171, 1969222, 1969253, 1969464, 1976997, 1985183, 1986028, 2000966, 2001989, 2028270, 2042460, 2052415, 2058421, 2059202, 2060522, 2069221, 2076393, 2130272, 2182162
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1439416 is a term because 1439416 = 1^5 + 1^5 + 2^5 + 7^5 + 8^5 + 8^5 + 10^5 + 12^5 + 12^5 + 15^5 = 1^5 + 2^5 + 3^5 + 6^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 3^5 + 6^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 1^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 8^5 + 9^5 + 12^5 + 16^5 = 2^5 + 2^5 + 3^5 + 4^5 + 6^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 + 15^5 = 3^5 + 3^5 + 3^5 + 3^5 + 6^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 7^5 + 11^5 + 11^5 + 16^5 = 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5.

Links

Crossrefs

Cf. A345603, A345627, A345641, A346355.

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, 10):
    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 1-5 of 5 results.

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