A345825 -id:A345825 - cp's OEIS frontend

A345569 Numbers that are the sum of seven fourth powers in three or more ways.

2677, 2692, 2757, 2852, 2867, 2917, 2932, 2997, 3107, 3172, 3301, 3476, 3541, 3972, 4132, 4147, 4212, 4227, 4242, 4257, 4307, 4322, 4372, 4387, 4437, 4452, 4482, 4497, 4562, 4627, 4737, 4756, 4851, 4866, 4867, 4931, 4996, 5077, 5106, 5107, 5122, 5187, 5252

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			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. A345521, A345560, A345568, A345570, A345578, A345606, A345825.

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])

A345815 Numbers that are the sum of six fourth powers in exactly three ways.

Original entry on oeis.org

2676, 2851, 2916, 4131, 4226, 4241, 4306, 4371, 4481, 4850, 5346, 5411, 5521, 5586, 5651, 6561, 6611, 6756, 6771, 6801, 6821, 6836, 6851, 6931, 7106, 7235, 7475, 7491, 7666, 7841, 7906, 7971, 8146, 8211, 8321, 8386, 8451, 8531, 8706, 9011, 9156, 9171, 9186

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345560 at term 18 because 6626 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.

			2851 is a term because 2851 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

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

Cf. A048931, A344244, A345560, A345814, A345816, A345825, A346358.

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, 6):
    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])

A345824 Numbers that are the sum of seven fourth powers in exactly two ways.

Original entry on oeis.org

262, 277, 292, 307, 342, 357, 372, 422, 437, 502, 517, 532, 547, 597, 612, 677, 772, 787, 852, 886, 901, 916, 966, 981, 1027, 1046, 1141, 1156, 1221, 1362, 1377, 1396, 1442, 1510, 1525, 1557, 1572, 1587, 1590, 1617, 1637, 1652, 1717, 1765, 1812, 1827, 1892

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345568 at term 61.

			277 is a term because 277 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 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. A345568, A345774, A345814, A345823, A345825, A345834, A346279.

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

A345826 Numbers that are the sum of seven fourth powers in exactly four ways.

Original entry on oeis.org

2932, 4147, 4212, 4387, 5427, 5602, 5667, 6627, 6692, 6817, 6822, 6837, 6852, 6867, 7012, 7122, 7251, 7316, 7491, 7747, 7857, 8052, 8097, 8162, 8402, 8467, 8532, 8707, 8787, 9027, 9092, 9157, 9172, 9202, 9237, 9252, 9332, 9412, 9442, 9492, 9572, 9652, 9682

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345570 at term 9 because 6642 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

			4147 is a term because 4147 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4.

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

Cf. A345570, A345776, A345816, A345825, A345827, A345836, A346281.

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

A345835 Numbers that are the sum of eight fourth powers in exactly three ways.

Original entry on oeis.org

518, 2678, 2693, 2708, 2738, 2758, 2773, 2838, 2853, 2868, 2883, 2918, 2998, 3078, 3108, 3123, 3253, 3302, 3317, 3363, 3382, 3428, 3477, 3492, 3542, 3622, 3732, 3778, 3797, 3893, 3926, 3953, 3973, 3988, 4018, 4053, 4101, 4118, 4133, 4166, 4193, 4243, 4258

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345578 at term 13 because 2933 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

			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. A345578, A345785, A345825, A345834, A345836, A345845, A346328.

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])

A345775 Numbers that are the sum of seven cubes in exactly three ways.

Original entry on oeis.org

222, 229, 248, 255, 262, 281, 283, 285, 318, 346, 370, 374, 377, 379, 381, 396, 400, 407, 412, 419, 426, 433, 437, 438, 444, 451, 463, 472, 475, 477, 489, 494, 501, 505, 507, 510, 522, 529, 533, 536, 559, 564, 566, 568, 570, 577, 578, 584, 585, 592, 594, 596

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345521 at term 28 because 470 = 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 = 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3.

Likely finite.

			229 is a term because 229 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 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..390

Cf. A048931, A345521, A345774, A345776, A345785, A345825.

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

A346280 Numbers that are the sum of seven fifth powers in exactly three ways.

Original entry on oeis.org

84457, 166997, 324860, 326199, 358482, 359327, 391007, 391999, 408158, 455146, 455749, 486468, 502429, 572054, 595519, 614505, 622280, 648319, 671210, 672022, 696468, 696499, 696710, 697491, 699592, 704243, 713274, 729235, 755516, 796467, 857518, 877645

Offset: 1

David Consiglio, Jr., Jul 13 2021

nonn

Differs from A345606 at term 39 because 893604 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 15^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5.

			84457 is a term because 84457 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 9^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5.

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

Cf. A345606, A345825, A346279, A346281, A346328, A346358.

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, 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])

cp's OEIS Frontend

A345569 Numbers that are the sum of seven fourth powers in three or more ways.

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A345815 Numbers that are the sum of six fourth powers in exactly three ways.

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A345824 Numbers that are the sum of seven fourth powers in exactly two ways.

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A345826 Numbers that are the sum of seven fourth powers in exactly four ways.

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A345835 Numbers that are the sum of eight fourth powers in exactly three ways.

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A345775 Numbers that are the sum of seven cubes in exactly three ways.

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A346280 Numbers that are the sum of seven fifth powers in exactly three ways.

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Python