A345594 -id:A345594 - cp's OEIS frontend

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

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

David Consiglio, Jr., Jun 20 2021

nonn

			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.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345548, A345584, A345592, A345594, A345602, A345626, A345851.

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])

A345549 Numbers that are the sum of nine cubes in ten or more ways.

Original entry on oeis.org

966, 971, 978, 985, 992, 1004, 1011, 1018, 1022, 1048, 1055, 1056, 1062, 1063, 1074, 1076, 1078, 1081, 1083, 1085, 1088, 1092, 1093, 1095, 1097, 1098, 1100, 1102, 1104, 1107, 1109, 1111, 1112, 1114, 1117, 1118, 1119, 1121, 1123, 1124, 1126, 1128, 1130, 1133

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			971 is a term because 971 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345540, A345548, A345558, A345594, A345802, A346803.

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, 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])

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

David Consiglio, Jr., Jun 20 2021

nonn

			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.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345540, A345576, A345584, A345594, A345618, A345842.

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])

A345852 Numbers that are the sum of nine fourth powers in exactly ten ways.

Original entry on oeis.org

9299, 12708, 12948, 13269, 13349, 13524, 13589, 13764, 13829, 13893, 14133, 14228, 14468, 14564, 14869, 14934, 14964, 15014, 15094, 15109, 15174, 15189, 15333, 15428, 15429, 15524, 15588, 15604, 15653, 16214, 16229, 16469, 16564, 16644, 16773, 16883, 16948

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345594 at term 15 because 14804 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 + 8^4 + 9^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 8^4 + 9^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 11^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 8^4 + 8^4 = 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 8^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 9^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 8^4 + 9^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 8^4 + 9^4 = 2^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 + 9^4 = 2^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4 + 7^4 + 7^4 + 7^4 = 3^4 + 4^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 + 9^4.

			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.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345594, A345802, A345842, A345851, A345862, A346345.

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])

A345603 Numbers that are the sum of ten fourth powers in ten or more ways.

Original entry on oeis.org

6885, 7990, 8035, 8100, 8165, 8275, 8340, 8515, 8565, 8580, 9140, 9205, 9235, 9285, 9300, 9315, 9380, 9445, 9495, 9510, 9540, 9555, 9620, 9670, 9685, 9750, 9795, 9830, 9860, 9924, 9925, 9990, 10005, 10164, 10294, 10340, 10374, 10404, 10420, 10515, 10534

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			7990 is a term because 7990 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 1^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 9^4 = 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345558, A345594, A345602, A345642, A345862.

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, 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])

A345627 Numbers that are the sum of nine fifth powers in ten or more ways.

Original entry on oeis.org

4157156, 4492410, 4510461, 4915538, 4948274, 5005474, 5015506, 5179747, 5219655, 5252477, 5739988, 5756794, 6323426, 6326519, 6382443, 6423394, 6654999, 6705284, 6793170, 6861218, 7101038, 7147645, 7147656, 7148679, 7266240, 7280391, 7283268, 7314187, 7413493

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			4492410 is a term because 4492410 = 1^5 + 1^5 + 2^5 + 3^5 + 5^5 + 7^5 + 7^5 + 13^5 + 21^5 = 1^5 + 2^5 + 6^5 + 10^5 + 11^5 + 11^5 + 14^5 + 16^5 + 19^5 = 1^5 + 6^5 + 7^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 + 18^5 = 2^5 + 5^5 + 6^5 + 6^5 + 7^5 + 15^5 + 15^5 + 16^5 + 18^5 = 2^5 + 5^5 + 6^5 + 10^5 + 10^5 + 11^5 + 11^5 + 15^5 + 20^5 = 3^5 + 3^5 + 7^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 + 18^5 = 3^5 + 3^5 + 8^5 + 8^5 + 8^5 + 12^5 + 12^5 + 17^5 + 19^5 = 3^5 + 4^5 + 6^5 + 7^5 + 8^5 + 13^5 + 14^5 + 16^5 + 19^5 = 4^5 + 4^5 + 4^5 + 7^5 + 11^5 + 11^5 + 13^5 + 18^5 + 18^5 = 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 9^5 + 16^5 + 17^5 + 18^5.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345594, A345618, A345626, A345642, A346345.

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])

cp's OEIS Frontend

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

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A345549 Numbers that are the sum of nine cubes in ten or more ways.

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

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A345852 Numbers that are the sum of nine fourth powers in exactly ten ways.

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A345603 Numbers that are the sum of ten fourth powers in ten or more ways.

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A345627 Numbers that are the sum of nine fifth powers in ten or more ways.

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