A345516 -id:A345516 - cp's OEIS frontend

A345180 Numbers that are the sum of five third powers in seven or more ways.

4392, 4472, 4544, 4600, 4915, 4957, 5076, 5113, 5120, 5132, 5139, 5165, 5174, 5256, 5321, 5347, 5354, 5384, 5391, 5410, 5445, 5474, 5481, 5507, 5543, 5617, 5624, 5643, 5678, 5715, 5741, 5760, 5769, 5797, 5832, 5834, 5860, 5895, 5914, 5923, 5984, 5986, 6049

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

			4472 is a term because 4472 = 1^3 + 4^3 + 4^3 + 4^3 + 15^3  = 2^3 + 2^3 + 9^3 + 11^3 + 11^3  = 2^3 + 3^3 + 4^3 + 5^3 + 15^3  = 2^3 + 3^3 + 7^3 + 11^3 + 12^3  = 3^3 + 3^3 + 6^3 + 10^3 + 13^3  = 3^3 + 4^3 + 5^3 + 8^3 + 14^3  = 5^3 + 5^3 + 7^3 + 10^3 + 12^3.

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Cf. A344800, A344942, A345150, A345174, A345181, A345183, A345516.

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, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 7])
for x in range(len(rets)):
    print(rets[x])

A345515 Numbers that are the sum of six cubes in six or more ways.

Original entry on oeis.org

1377, 1488, 1586, 1595, 1647, 1673, 1677, 1710, 1738, 1764, 1766, 1773, 1799, 1829, 1836, 1837, 1862, 1881, 1890, 1911, 1953, 1955, 1981, 1988, 2007, 2011, 2014, 2018, 2025, 2044, 2051, 2070, 2079, 2097, 2105, 2107, 2108, 2142, 2153, 2160, 2168, 2170, 2177

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			1488 is a term because 1488 = 1^3 + 1^3 + 1^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.

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

Cf. A344810, A345174, A345514, A345516, A345524, A345563, A345768.

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, 6):
    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])

A345517 Numbers that are the sum of six cubes in eight or more ways.

Original entry on oeis.org

1981, 2105, 2168, 2277, 2368, 2376, 2431, 2438, 2457, 2466, 2494, 2538, 2555, 2557, 2583, 2593, 2646, 2665, 2672, 2709, 2746, 2753, 2763, 2765, 2772, 2880, 2881, 2889, 2916, 2942, 2961, 2970, 2977, 2979, 2980, 2987, 3007, 3033, 3040, 3042, 3043, 3049, 3068

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			2105 is a term because 2105 = 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 11^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 11^3 = 1^3 + 2^3 + 6^3 + 7^3 + 7^3 + 8^3 = 1^3 + 4^3 + 4^3 + 4^3 + 8^3 + 9^3 = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 10^3 = 2^3 + 3^3 + 4^3 + 5^3 + 8^3 + 9^3 = 3^3 + 3^3 + 3^3 + 7^3 + 7^3 + 9^3 = 5^3 + 5^3 + 5^3 + 5^3 + 7^3 + 8^3.

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

Cf. A344812, A345183, A345516, A345518, A345526, A345565, A345770.

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, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 8])
    for x in range(len(rets)):
        print(rets[x])

A345525 Numbers that are the sum of seven cubes in seven or more ways.

Original entry on oeis.org

1072, 1170, 1235, 1261, 1268, 1305, 1385, 1392, 1396, 1411, 1440, 1441, 1448, 1450, 1459, 1489, 1496, 1502, 1504, 1513, 1515, 1538, 1540, 1547, 1552, 1557, 1559, 1564, 1565, 1566, 1567, 1576, 1585, 1587, 1592, 1593, 1594, 1600, 1602, 1603, 1606, 1613, 1620

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			1170 is a term because 1170 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3 = 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 7^3.

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

Cf. A345484, A345516, A345524, A345526, A345537, A345573, A345779.

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 >= 7])
    for x in range(len(rets)):
        print(rets[x])

A345564 Numbers that are the sum of six fourth powers in seven or more ways.

Original entry on oeis.org

21251, 43875, 48276, 49796, 53315, 58035, 58500, 59780, 59795, 59811, 67875, 68306, 69155, 69779, 71955, 72051, 72131, 73970, 74420, 74851, 77010, 80291, 80515, 81875, 82275, 84515, 86436, 86451, 86531, 87075, 87746, 88355, 88595, 88660, 88675, 90355, 91475

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			43875 is a term because 43875 = 1^4 + 2^4 + 9^4 + 9^4 + 10^4 + 12^4 = 2^4 + 2^4 + 2^4 + 5^4 + 11^4 + 13^4 = 2^4 + 2^4 + 5^4 + 7^4 + 7^4 + 14^4 = 2^4 + 5^4 + 6^4 + 9^4 + 11^4 + 12^4 = 3^4 + 7^4 + 8^4 + 9^4 + 10^4 + 12^4 = 4^4 + 4^4 + 7^4 + 7^4 + 10^4 + 13^4 = 5^4 + 7^4 + 8^4 + 8^4 + 8^4 + 13^4.

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

Cf. A344942, A345516, A345563, A345565, A345573, A345721, A345819.

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 >= 7])
    for x in range(len(rets)):
        print(rets[x])

A345769 Numbers that are the sum of six cubes in exactly seven ways.

Original entry on oeis.org

1710, 1766, 1773, 1988, 2051, 2160, 2196, 2249, 2251, 2259, 2314, 2322, 2349, 2375, 2417, 2424, 2480, 2492, 2513, 2520, 2531, 2539, 2548, 2564, 2565, 2574, 2611, 2613, 2639, 2656, 2702, 2707, 2762, 2770, 2773, 2792, 2798, 2808, 2818, 2825, 2826, 2833, 2844

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345516 at term 4 because 1981 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 12^3 = 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 12^3 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 10^3 = 1^3 + 1^3 + 6^3 + 6^3 + 6^3 + 11^3 = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 10^3 = 3^3 + 3^3 + 7^3 + 7^3 + 8^3 + 9^3 = 3^3 + 4^3 + 6^3 + 6^3 + 9^3 + 9^3 = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 10^3.

			1766 is a term because 1766 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 11^3 = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 9^3 = 1^3 + 3^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 9^3 = 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 9^3 = 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3.

Sean A. Irvine, Table of n, a(n) for n = 1..1359

Cf. A345181, A345516, A345768, A345770, A345779, A345819.

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, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 7])
    for x in range(len(rets)):
        print(rets[x])

A344811 Numbers that are the sum of six squares in seven or more ways.

Original entry on oeis.org

60, 65, 68, 69, 77, 78, 81, 84, 86, 87, 89, 90, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			65 = 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 3^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 6^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 6^2
   = 2^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2
   = 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2
so 65 is a term.

Sean A. Irvine, Table of n, a(n) for n = 1..1000

Cf. A344800, A344810, A344812, A345484, A345516.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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 >= 7])
    for x in range(len(rets)):
        print(rets[x])

cp's OEIS Frontend

A345180 Numbers that are the sum of five third powers in seven or more ways.

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A345515 Numbers that are the sum of six cubes in six or more ways.

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A345517 Numbers that are the sum of six cubes in eight or more ways.

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A345525 Numbers that are the sum of seven cubes in seven or more ways.

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A345564 Numbers that are the sum of six fourth powers in seven or more ways.

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A345769 Numbers that are the sum of six cubes in exactly seven ways.

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A344811 Numbers that are the sum of six squares in seven or more ways.

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Python