A345783 -id:A345783 - cp's OEIS frontend

A345784 Numbers that are the sum of eight cubes in exactly two ways.

132, 139, 158, 160, 167, 174, 181, 186, 193, 195, 197, 200, 212, 216, 219, 238, 244, 251, 258, 265, 272, 277, 288, 296, 298, 300, 301, 303, 307, 314, 315, 317, 321, 322, 327, 328, 329, 333, 334, 336, 338, 340, 341, 348, 350, 352, 356, 359, 360, 361, 363, 366

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345532 at term 16 because 223 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3.

Likely finite.

			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..173

Cf. A345532, A345774, A345783, A345785, A345794, A345834.

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])

A345833 Numbers that are the sum of eight fourth powers in exactly one ways.

Original entry on oeis.org

8, 23, 38, 53, 68, 83, 88, 98, 103, 113, 118, 128, 133, 148, 163, 168, 178, 183, 193, 198, 213, 228, 243, 248, 258, 328, 338, 353, 368, 403, 408, 418, 433, 468, 483, 488, 498, 568, 578, 593, 608, 632, 643, 647, 648, 658, 662, 663, 673, 677, 692, 707, 708, 712

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A003342 at term 26 because 263 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4.

			23 is a term because 23 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4.

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

Cf. A003342, A345783, A345823, A345834, A345843, A346326.

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 == 1])
    for x in range(len(rets)):
        print(rets[x])

A345793 Numbers that are the sum of nine cubes in exactly one way.

Original entry on oeis.org

9, 16, 23, 30, 35, 37, 42, 44, 49, 51, 56, 58, 61, 63, 65, 68, 70, 75, 77, 79, 82, 84, 86, 87, 89, 91, 93, 94, 96, 98, 100, 101, 103, 105, 107, 108, 110, 112, 113, 114, 115, 119, 120, 121, 122, 124, 126, 127, 128, 129, 131, 134, 135, 138, 139, 141, 142, 145

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A003332 at term 18 because 72 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^2 + 1^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3.

Likely finite.

			16 is a term because 16 = 1^3 + 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..176

Cf. A003332, A345783, A345794, A345803, A345843.

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 == 1])
    for x in range(len(rets)):
        print(rets[x])

A345773 Numbers that are the sum of seven cubes in exactly one way.

Original entry on oeis.org

7, 14, 21, 28, 33, 35, 40, 42, 47, 49, 54, 56, 59, 61, 66, 68, 70, 73, 75, 77, 80, 84, 85, 87, 91, 92, 94, 96, 98, 99, 103, 105, 106, 110, 111, 112, 113, 117, 118, 122, 124, 125, 129, 132, 133, 136, 137, 138, 140, 143, 144, 145, 147, 148, 150, 151, 152, 154

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A003330 at term 44 because 131 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.

Likely finite.

			14 is a term because 14 = 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..324

Cf. A003330, A048929, A345774, A345783, A345823.

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 == 1])
    for x in range(len(rets)):
        print(rets[x])

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A345784 Numbers that are the sum of eight cubes in exactly two ways.

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A345833 Numbers that are the sum of eight fourth powers in exactly one ways.

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A345793 Numbers that are the sum of nine cubes in exactly one way.

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A345773 Numbers that are the sum of seven cubes in exactly one way.

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Python