A345835 -id:A345835 - cp's OEIS frontend

A345578 Numbers that are the sum of eight fourth powers in three or more ways.

518, 2678, 2693, 2708, 2738, 2758, 2773, 2838, 2853, 2868, 2883, 2918, 2933, 2948, 2998, 3013, 3078, 3108, 3123, 3173, 3188, 3253, 3302, 3317, 3363, 3382, 3428, 3477, 3492, 3542, 3557, 3622, 3732, 3778, 3797, 3893, 3926, 3953, 3973, 3988, 4018, 4053, 4101

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			2678 is a term because 2678 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

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

Cf. A345533, A345569, A345577, A345579, A345587, A345611, A345835.

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

A345825 Numbers that are the sum of seven fourth powers in exactly three ways.

Original entry on oeis.org

2677, 2692, 2757, 2852, 2867, 2917, 2997, 3107, 3172, 3301, 3476, 3541, 3972, 4132, 4227, 4242, 4257, 4307, 4322, 4372, 4437, 4452, 4482, 4497, 4562, 4627, 4737, 4756, 4851, 4866, 4867, 4931, 4996, 5077, 5106, 5107, 5122, 5187, 5252, 5282, 5317, 5347, 5362

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345569 at term 7 because 2932 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

			2692 is a term because 2692 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

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

Cf. A345569, A345775, A345815, A345824, A345826, A345835, A346280.

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, 7):
    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])

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

Original entry on oeis.org

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])

A345836 Numbers that are the sum of eight fourth powers in exactly four ways.

Original entry on oeis.org

2933, 2948, 3013, 3173, 3188, 3557, 4148, 4163, 4213, 4293, 4388, 4453, 4643, 4772, 4837, 4883, 5012, 5123, 5188, 5203, 5268, 5333, 5363, 5378, 5398, 5428, 5538, 5573, 5603, 5618, 5668, 5733, 5748, 5858, 5923, 6052, 6163, 6227, 6292, 6548, 6578, 6628, 6693

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345579 at term 10 because 4228 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4.

			2948 is a term because 2948 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 2^4 + 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. A345579, A345786, A345826, A345835, A345837, A345846, A346329.

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

A345845 Numbers that are the sum of nine fourth powers in exactly three ways.

Original entry on oeis.org

519, 534, 599, 774, 1143, 1364, 1539, 1604, 1619, 1814, 2579, 2644, 2659, 2679, 2694, 2709, 2724, 2739, 2754, 2759, 2774, 2789, 2819, 2834, 2839, 2869, 2884, 2899, 2994, 2999, 3079, 3109, 3124, 3139, 3303, 3318, 3333, 3334, 3363, 3364, 3379, 3383, 3398, 3463

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345587 at term 26 because 285.

			534 is a term because 534 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 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. A345587, A345795, A345835, A345844, A345846, A345855, A346338.

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 == 3])
    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])

A346328 Numbers that are the sum of eight fifth powers in exactly three ways.

Original entry on oeis.org

52417, 54518, 69634, 70954, 84458, 84489, 84700, 85481, 87582, 92233, 101264, 102890, 112574, 117225, 119326, 134473, 143264, 143442, 143506, 149781, 151448, 158719, 159465, 165634, 166998, 167029, 167196, 167240, 168021, 170122, 174773, 183804, 184457

Offset: 1

David Consiglio, Jr., Jul 13 2021

nonn

Differs from A345611 at term 105 because 391250 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5.

			52417 is a term because 52417 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.

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

Cf. A345611, A345835, A346280, A346327, A346329, A346338.

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, 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])

cp's OEIS Frontend

A345578 Numbers that are the sum of eight fourth powers in three or more ways.

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A345825 Numbers that are the sum of seven fourth powers in exactly three ways.

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

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A345836 Numbers that are the sum of eight fourth powers in exactly four ways.

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A345845 Numbers that are the sum of nine fourth powers in exactly three ways.

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

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A346328 Numbers that are the sum of eight fifth powers in exactly three ways.

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Python