A345595 -id:A345595 - cp's OEIS frontend

A003344 Numbers that are the sum of 10 positive 4th powers.

10, 25, 40, 55, 70, 85, 90, 100, 105, 115, 120, 130, 135, 145, 150, 160, 165, 170, 180, 185, 195, 200, 210, 215, 225, 230, 245, 250, 260, 265, 275, 280, 290, 295, 310, 325, 330, 340, 345, 355, 360, 370, 375, 385, 390, 400, 405, 410, 420, 425, 435, 440, 450, 455, 465

Offset: 1

N. J. A. Sloane

			From _David A. Corneth_, Aug 03 2020: (Start)
5176 is in the sequence as 5176 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 5^4 + 5^4 + 5^4 + 5^4 + 7^4.
6901 is in the sequence as 6901 = 1^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4.
8502 is in the sequence as 8502 = 1^4 + 3^4 + 4^4 + 5^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4 + 7^4. (End)

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Eric Weisstein's World of Mathematics, Biquadratic Number.

Cf. A000583, A003333, A003343, A003355, A047721, A345595, A345853.

from itertools import count, takewhile, combinations_with_replacement as mc
def aupto(limit):
    pows4 = list(takewhile(lambda x: x <= limit, (i**4 for i in count(1))))
    sum10 = set(sum(c) for c in mc(pows4, 10) if sum(c) <= limit)
    return sorted(sum10)
print(aupto(465)) # Michael S. Branicky, Oct 25 2021

A345586 Numbers that are the sum of nine fourth powers in two or more ways.

Original entry on oeis.org

264, 279, 294, 309, 324, 339, 344, 359, 374, 389, 404, 424, 439, 454, 469, 504, 519, 534, 549, 564, 579, 584, 599, 614, 629, 644, 664, 679, 694, 709, 759, 774, 789, 804, 819, 839, 854, 869, 884, 888, 903, 918, 933, 934, 948, 949, 968, 983, 998, 1013, 1014

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			279 is a term because 279 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

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

Cf. A003343, A345541, A345577, A345587, A345595, A345619, A345844.

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 >= 2])
    for x in range(len(rets)):
        print(rets[x])

A345596 Numbers that are the sum of ten fourth powers in three or more ways.

Original entry on oeis.org

520, 535, 550, 600, 615, 680, 775, 790, 855, 1030, 1144, 1159, 1224, 1365, 1380, 1399, 1445, 1540, 1555, 1605, 1620, 1635, 1685, 1700, 1768, 1795, 1815, 1830, 1860, 1875, 1895, 1989, 2070, 2164, 2229, 2244, 2439, 2485, 2580, 2595, 2645, 2660, 2675, 2680, 2695

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			535 is a term because 535 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

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

Cf. A345551, A345587, A345595, A345597, A345635, A345855.

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 >= 3])
    for x in range(len(rets)):
        print(rets[x])

A345854 Numbers that are the sum of ten fourth powers in exactly two ways.

Original entry on oeis.org

265, 280, 295, 310, 325, 340, 345, 355, 360, 375, 390, 405, 420, 425, 440, 455, 470, 485, 505, 565, 580, 585, 595, 630, 645, 660, 665, 695, 710, 725, 745, 760, 805, 820, 835, 840, 870, 885, 889, 900, 904, 919, 920, 934, 935, 949, 950, 964, 965, 969, 984, 999

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345595 at term 20 because 520 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

			280 is a term because 280 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

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

Cf. A345595, A345804, A345844, A345853, A345855, A346347.

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 == 2])
    for x in range(len(rets)):
        print(rets[x])

A345550 Numbers that are the sum of ten cubes in two or more ways.

Original entry on oeis.org

73, 80, 99, 134, 136, 141, 148, 155, 160, 162, 167, 169, 174, 176, 183, 186, 188, 190, 192, 193, 195, 197, 199, 202, 204, 206, 209, 211, 212, 213, 214, 216, 218, 221, 223, 225, 228, 230, 232, 235, 239, 240, 244, 246, 247, 249, 251, 253, 254, 258, 260, 262

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			80 is a term because 80 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3.

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

Cf. A003333, A345509, A345541, A345551, A345595, A345804.

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, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 2])
    for x in range(len(rets)):
        print(rets[x])

A345634 Numbers that are the sum of ten fifth powers in two or more ways.

Original entry on oeis.org

4102, 4133, 4164, 4195, 4226, 4257, 4344, 4375, 4406, 4437, 4468, 4586, 4617, 4648, 4679, 4828, 4859, 4890, 5070, 5101, 5125, 5156, 5187, 5218, 5249, 5312, 5367, 5398, 5429, 5460, 5609, 5640, 5671, 5851, 5882, 6093, 6148, 6179, 6210, 6241, 6390, 6421, 6452

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			4133 is a term because 4133 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 = 1^5 + 1^5 + 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5.

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

Cf. A003355, A345595, A345619, A345635, A346347.

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 >= 2])
    for x in range(len(rets)):
        print(rets[x])

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A003344 Numbers that are the sum of 10 positive 4th powers.

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A345586 Numbers that are the sum of nine fourth powers in two or more ways.

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A345596 Numbers that are the sum of ten fourth powers in three or more ways.

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A345854 Numbers that are the sum of ten fourth powers in exactly two ways.

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A345550 Numbers that are the sum of ten cubes in two or more ways.

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A345634 Numbers that are the sum of ten fifth powers in two or more ways.

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