A345784 -id:A345784 - cp's OEIS frontend

A345834 Numbers that are the sum of eight fourth powers in exactly two ways.

263, 278, 293, 308, 323, 343, 358, 373, 388, 423, 438, 453, 503, 533, 548, 563, 583, 598, 613, 628, 678, 693, 758, 773, 788, 803, 853, 868, 887, 902, 917, 932, 933, 967, 982, 997, 1028, 1043, 1047, 1062, 1108, 1127, 1142, 1157, 1172, 1222, 1237, 1283, 1302

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345577 at term 14 because 518 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

			278 is a term because 278 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 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. A345577, A345784, A345824, A345833, A345835, A345844, A346327.

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

A345532 Numbers that are the sum of eight cubes in two or more ways.

Original entry on oeis.org

132, 139, 158, 160, 167, 174, 181, 186, 193, 195, 197, 200, 212, 216, 219, 223, 230, 237, 238, 244, 249, 251, 256, 258, 263, 265, 270, 272, 275, 277, 282, 284, 286, 288, 289, 291, 293, 296, 298, 300, 301, 303, 307, 308, 310, 312, 314, 315, 317, 319, 321, 322

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			139 is a term because 139 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3.

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

Cf. A003331, A345489, A345520, A345533, A345541, A345577, A345784.

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, 8):
    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])

A345774 Numbers that are the sum of seven cubes in exactly two ways.

Original entry on oeis.org

131, 159, 166, 173, 185, 192, 211, 236, 243, 257, 264, 269, 274, 276, 288, 290, 292, 295, 299, 300, 302, 307, 309, 311, 314, 320, 321, 325, 332, 333, 337, 339, 340, 344, 348, 351, 353, 355, 358, 359, 360, 363, 372, 384, 385, 386, 388, 389, 393, 395, 398, 403

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345520 at term 8 because 222 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3.

Likely finite.

			159 is a term because 159 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3.

Sean A. Irvine, Table of n, a(n) for n = 1..355

Cf. A048930, A345520, A345773, A345775, A345784, A345824.

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, 7):
    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])

A345785 Numbers that are the sum of eight cubes in exactly three ways.

Original entry on oeis.org

223, 230, 237, 249, 263, 270, 275, 282, 284, 286, 289, 291, 293, 308, 310, 312, 319, 326, 345, 349, 354, 364, 371, 373, 375, 378, 380, 385, 386, 387, 389, 397, 399, 404, 406, 410, 412, 413, 415, 420, 423, 439, 441, 443, 446, 449, 452, 453, 459, 460, 465, 473

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345533 at term 5 because 256 = 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 + 2^3 + 3^3 + 6^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3.

Likely finite.

			230 is a term because 230 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Sean A. Irvine, Table of n, a(n) for n = 1..198

Cf. A345533, A345775, A345784, A345786, A345795, A345835.

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, 8):
    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])

A345794 Numbers that are the sum of nine cubes in exactly two ways.

Original entry on oeis.org

72, 133, 140, 147, 159, 161, 166, 168, 175, 182, 185, 187, 189, 194, 196, 198, 201, 203, 205, 208, 213, 217, 220, 222, 227, 239, 243, 246, 252, 261, 265, 266, 273, 289, 296, 304, 306, 308, 323, 325, 328, 329, 330, 336, 342, 344, 349, 351, 352, 354, 356, 358

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345541 at term 25 because 224 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

Likely finite.

			133 is a term because 133 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3.

Sean A. Irvine, Table of n, a(n) for n = 1..105

Cf. A345541, A345784, A345793, A345795, A345804, A345844.

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

A345783 Numbers that are the sum of eight cubes in exactly one way.

Original entry on oeis.org

8, 15, 22, 29, 34, 36, 41, 43, 48, 50, 55, 57, 60, 62, 64, 67, 69, 71, 74, 76, 78, 81, 83, 85, 86, 88, 92, 93, 95, 97, 99, 100, 102, 104, 106, 107, 111, 112, 113, 114, 118, 119, 120, 121, 123, 125, 126, 130, 133, 134, 137, 138, 140, 141, 144, 145, 146, 148

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A003331 at term 49 because 132 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Likely finite.

			15 is a term because 15 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3.

Sean A. Irvine, Table of n, a(n) for n = 1..209

Cf. A003331, A345773, A345784, A345793, A345833.

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, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 1])
    for x in range(len(rets)):
        print(rets[x])

cp's OEIS Frontend

A345834 Numbers that are the sum of eight fourth powers in exactly two ways.

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A345532 Numbers that are the sum of eight cubes in two or more ways.

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A345774 Numbers that are the sum of seven cubes in exactly two ways.

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A345785 Numbers that are the sum of eight cubes in exactly three ways.

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A345794 Numbers that are the sum of nine cubes in exactly two ways.

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A345783 Numbers that are the sum of eight cubes in exactly one way.

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