A345847 -id:A345847 - cp's OEIS frontend

A345589 Numbers that are the sum of nine fourth powers in five or more ways.

3189, 4149, 4229, 4244, 4309, 4374, 4404, 4419, 4469, 4484, 4549, 4659, 4724, 4853, 4899, 5028, 5093, 5139, 5189, 5204, 5269, 5284, 5349, 5379, 5414, 5444, 5459, 5509, 5524, 5574, 5589, 5619, 5634, 5654, 5684, 5699, 5749, 5764, 5814, 5829, 5939, 6068, 6133

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			4149 is a term because 4149 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 = 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4.

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

Cf. A345544, A345580, A345588, A345590, A345598, A345622, A345847.

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])

A345837 Numbers that are the sum of eight fourth powers in exactly five ways.

Original entry on oeis.org

4228, 4403, 4468, 5443, 5508, 5683, 6613, 6643, 6658, 6708, 6773, 6838, 6868, 6883, 6948, 7013, 7093, 7138, 7203, 7267, 7268, 7332, 7397, 7478, 7507, 7572, 7588, 7828, 7858, 7923, 7988, 8113, 8133, 8228, 8353, 8418, 8533, 8547, 8548, 8612, 8723, 8788, 8852

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345580 at term 11 because 6723 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

			4403 is a term because 4403 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.

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

Cf. A345580, A345787, A345827, A345836, A345838, A345847, A346330.

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 == 5])
    for x in range(len(rets)):
        print(rets[x])

A345846 Numbers that are the sum of nine fourth powers in exactly four ways.

Original entry on oeis.org

2854, 2919, 2934, 2949, 2964, 3014, 3029, 3094, 3159, 3174, 3204, 3254, 3269, 3429, 3444, 3558, 3573, 3638, 3798, 3813, 3974, 4034, 4134, 4164, 4179, 4182, 4209, 4214, 4274, 4294, 4389, 4439, 4454, 4534, 4614, 4644, 4709, 4773, 4788, 4838, 4884, 4918, 4949

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345588 at term 11 because 3189 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4.

			2919 is a term because 2919 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4.

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

Cf. A345588, A345796, A345836, A345845, A345847, A345856, A346339.

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 == 4])
    for x in range(len(rets)):
        print(rets[x])

A345848 Numbers that are the sum of nine fourth powers in exactly six ways.

Original entry on oeis.org

4469, 4484, 5444, 5459, 5524, 5589, 5699, 5764, 6629, 6659, 6674, 6694, 6724, 6789, 6884, 6899, 6914, 6934, 6949, 6964, 7014, 7154, 7219, 7334, 7348, 7349, 7413, 7459, 7478, 7494, 7523, 7524, 7588, 7589, 7604, 7653, 7669, 7734, 7779, 7874, 7954, 7989, 8069

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345590 at term 14 because 6739 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

			4484 is a term because 4484 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.

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

Cf. A345590, A345798, A345838, A345847, A345849, A345858, A346341.

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 == 6])
    for x in range(len(rets)):
        print(rets[x])

A345797 Numbers that are the sum of nine cubes in exactly five ways.

Original entry on oeis.org

409, 413, 428, 435, 439, 446, 465, 479, 491, 502, 512, 517, 526, 531, 533, 535, 538, 540, 559, 561, 563, 566, 568, 570, 576, 579, 580, 587, 594, 600, 601, 603, 613, 615, 617, 620, 622, 627, 632, 633, 635, 638, 646, 651, 653, 664, 665, 668, 670, 675, 680, 683

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345544 at term 8 because 472 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3.

Likely finite.

			413 is a term because 413 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.

Sean A. Irvine, Table of n, a(n) for n = 1..108

Cf. A345544, A345787, A345796, A345798, A345807, A345847.

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 == 5])
    for x in range(len(rets)):
        print(rets[x])

A345857 Numbers that are the sum of ten fourth powers in exactly five ways.

Original entry on oeis.org

2935, 3110, 3190, 3205, 3270, 3445, 3814, 3940, 4165, 4180, 4195, 4215, 4245, 4260, 4290, 4310, 4325, 4375, 4420, 4435, 4615, 4660, 4675, 4695, 4774, 4805, 4854, 4869, 4870, 4900, 4934, 4965, 4999, 5029, 5030, 5044, 5045, 5095, 5110, 5125, 5140, 5174, 5235

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345598 at term 3 because 3175 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

			3110 is a term because 3110 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

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

Cf. A345598, A345807, A345847, A345856, A345858, A346350.

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 == 5])
    for x in range(len(rets)):
        print(rets[x])

A346340 Numbers that are the sum of nine fifth powers in exactly five ways.

Original entry on oeis.org

392063, 392274, 406559, 458875, 519237, 538291, 607947, 663871, 672024, 672055, 672266, 672297, 673586, 673797, 674578, 675390, 680041, 681330, 704582, 704822, 714299, 730260, 732603, 763027, 763324, 765873, 766417, 777820, 780099, 814082, 820887, 825678

Offset: 1

David Consiglio, Jr., Jul 13 2021

nonn

Differs from A345622 at term 50 because 926404 = 2^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 10^5 + 15^5 = 2^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 10^5 + 15^5 = 2^5 + 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 2^5 + 7^5 + 7^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 2^5 + 6^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 1^5 + 4^5 + 4^5 + 7^5 + 11^5 + 12^5 + 12^5 + 12^5.

			392063 is a term because 392063 = 2^5 + 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 2^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 1^5 + 1^5 + 1^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5.

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

Cf. A345622, A345847, A346330, A346339, A346341, A346350.

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 == 5])
    for x in range(len(rets)):
        print(rets[x])

cp's OEIS Frontend

A345589 Numbers that are the sum of nine fourth powers in five or more ways.

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A345837 Numbers that are the sum of eight fourth powers in exactly five ways.

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A345846 Numbers that are the sum of nine fourth powers in exactly four ways.

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A345848 Numbers that are the sum of nine fourth powers in exactly six ways.

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A345797 Numbers that are the sum of nine cubes in exactly five ways.

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A345857 Numbers that are the sum of ten fourth powers in exactly five ways.

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A346340 Numbers that are the sum of nine fifth powers in exactly five ways.

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