A345807 -id:A345807 - cp's OEIS frontend

A345553 Numbers that are the sum of ten cubes in five or more ways.

288, 349, 382, 384, 401, 403, 408, 410, 414, 415, 417, 421, 429, 436, 440, 443, 447, 454, 455, 462, 466, 473, 475, 477, 480, 482, 487, 492, 496, 499, 501, 503, 506, 508, 510, 513, 515, 518, 520, 525, 527, 529, 532, 534, 536, 538, 539, 541, 543, 544, 545, 546

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			349 is a term because 349 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.

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

Cf. A345544, A345552, A345554, A345598, A345807, A346804.

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 >= 5])
    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])

A345806 Numbers that are the sum of ten cubes in exactly four ways.

Original entry on oeis.org

225, 232, 251, 258, 265, 272, 284, 286, 291, 307, 310, 314, 321, 323, 328, 342, 347, 356, 363, 366, 373, 375, 377, 380, 389, 391, 398, 399, 405, 412, 419, 422, 424, 427, 434, 438, 441, 445, 450, 451, 458, 459, 461, 464, 469, 471, 476, 478, 481, 484, 488, 489

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345552 at term 9 because 288 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.

Likely finite.

			232 is a term because 232 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

Sean A. Irvine, Table of n, a(n) for n = 1..93

Cf. A345552, A345796, A345805, A345807, A345856.

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

A345808 Numbers that are the sum of ten cubes in exactly six ways.

Original entry on oeis.org

436, 447, 466, 477, 480, 492, 503, 508, 510, 513, 515, 518, 527, 529, 536, 538, 539, 541, 551, 553, 560, 562, 564, 569, 577, 581, 590, 595, 601, 602, 603, 607, 608, 613, 614, 616, 618, 621, 628, 634, 636, 642, 643, 645, 647, 649, 654, 655, 659, 666, 678, 679

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345554 at term 2 because 440 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3.

Likely finite.

			440 is a term because 440 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1^3 + 1^3 + 3^3 + 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 + 3^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 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 + 3^3 + 5^3.

Sean A. Irvine, Table of n, a(n) for n = 1..73

Cf. A345554, A345798, A345807, A345809, A345858.

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

cp's OEIS Frontend

A345553 Numbers that are the sum of ten cubes in five or more 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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A345806 Numbers that are the sum of ten cubes in exactly four ways.

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A345808 Numbers that are the sum of ten cubes in exactly six ways.

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