A345820 -id:A345820 - cp's OEIS frontend

A345565 Numbers that are the sum of six fourth powers in eight or more ways.

58035, 59780, 87746, 88595, 96195, 96450, 102371, 106451, 106515, 108035, 108275, 108290, 108771, 112370, 112931, 115251, 122835, 122850, 122915, 124691, 125971, 132546, 133395, 133571, 133586, 134675, 134931, 136931, 138275, 138595, 143650, 144755, 144835

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			59780 is a term because 59780 = 1^4 + 1^4 + 1^4 + 5^4 + 12^4 + 14^4 = 1^4 + 1^4 + 6^4 + 6^4 + 9^4 + 15^4 = 1^4 + 2^4 + 9^4 + 10^4 + 11^4 + 13^4 = 1^4 + 4^4 + 7^4 + 7^4 + 8^4 + 15^4 = 1^4 + 7^4 + 7^4 + 9^4 + 10^4 + 14^4 = 2^4 + 5^4 + 6^4 + 11^4 + 11^4 + 13^4 = 3^4 + 7^4 + 8^4 + 10^4 + 11^4 + 13^4 = 5^4 + 6^4 + 7^4 + 7^4 + 11^4 + 14^4.

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

Cf. A344944, A345517, A345564, A345566, A345574, A345722, A345820.

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])

A345819 Numbers that are the sum of six fourth powers in exactly seven ways.

Original entry on oeis.org

21251, 43875, 48276, 49796, 53315, 58500, 59795, 59811, 67875, 68306, 69155, 69779, 71955, 72051, 72131, 73970, 74420, 74851, 77010, 80291, 80515, 81875, 82275, 84515, 86436, 86451, 86531, 87075, 88355, 88660, 88675, 90355, 91475, 93410, 93650, 94690, 95155

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345564 at term 6 because 58035 = 1^4 + 1^4 + 9^4 + 10^4 + 12^4 + 12^4 = 1^4 + 4^4 + 5^4 + 8^4 + 11^4 + 14^4 = 1^4 + 5^4 + 6^4 + 11^4 + 12^4 + 12^4 = 2^4 + 2^4 + 4^4 + 5^4 + 13^4 + 13^4 = 2^4 + 6^4 + 6^4 + 7^4 + 7^4 + 15^4 = 2^4 + 8^4 + 10^4 + 11^4 + 11^4 + 11^4 = 3^4 + 4^4 + 4^4 + 4^4 + 9^4 + 15^4 = 4^4 + 5^4 + 6^4 + 9^4 + 12^4 + 13^4.

			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. A344943, A345564, A345769, A345818, A345820, A345829, A346362.

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])

A345821 Numbers that are the sum of six fourth powers in exactly nine ways.

Original entry on oeis.org

88595, 132546, 134931, 144835, 146450, 162355, 170275, 171555, 171795, 172036, 172835, 177380, 177716, 180770, 183540, 184835, 185555, 187700, 187715, 190100, 190211, 193635, 195380, 195780, 196435, 197780, 199075, 199475, 199730, 199955, 202196, 202980

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345566 at term 2 because 122915 = 1^4 + 3^4 + 6^4 + 9^4 + 10^4 + 18^4 = 1^4 + 4^4 + 7^4 + 8^4 + 15^4 + 16^4 = 1^4 + 7^4 + 9^4 + 10^4 + 14^4 + 16^4 = 2^4 + 3^4 + 4^4 + 5^4 + 14^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 11^4 + 18^4 = 2^4 + 9^4 + 9^4 + 12^4 + 14^4 + 15^4 = 3^4 + 5^4 + 6^4 + 6^4 + 11^4 + 18^4 = 3^4 + 8^4 + 10^4 + 11^4 + 13^4 + 16^4 = 5^4 + 6^4 + 7^4 + 11^4 + 14^4 + 16^4 = 8^4 + 8^4 + 9^4 + 10^4 + 11^4 + 17^4.

			122915 is a term because 122915 = 1^4 + 3^4 + 6^4 + 9^4 + 10^4 + 18^4 = 1^4 + 4^4 + 7^4 + 8^4 + 15^4 + 16^4 = 1^4 + 7^4 + 9^4 + 10^4 + 14^4 + 16^4 = 2^4 + 3^4 + 4^4 + 5^4 + 14^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 11^4 + 18^4 = 2^4 + 9^4 + 9^4 + 12^4 + 14^4 + 15^4 = 3^4 + 5^4 + 6^4 + 6^4 + 11^4 + 18^4 = 3^4 + 8^4 + 10^4 + 11^4 + 13^4 + 16^4 = 5^4 + 6^4 + 7^4 + 11^4 + 14^4 + 16^4 = 8^4 + 8^4 + 9^4 + 10^4 + 11^4 + 17^4.

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

Cf. A341892, A345566, A345771, A345820, A345822, A345831, A346364.

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 == 9])
    for x in range(len(rets)):
        print(rets[x])

A345830 Numbers that are the sum of seven fourth powers in exactly eight ways.

Original entry on oeis.org

21252, 21507, 21636, 21876, 25347, 30372, 31412, 31652, 32116, 32356, 33811, 33907, 35637, 35652, 35892, 36261, 37827, 38052, 38596, 38676, 39267, 39347, 39971, 39972, 40212, 40452, 41506, 41731, 41987, 42147, 42227, 42357, 42532, 42771, 42852, 43027, 43282

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345574 at term 1 because 19491 = 1^4 + 1^4 + 1^4 + 6^4 + 8^4 + 8^4 + 10^4 = 1^4 + 2^4 + 4^4 + 4^4 + 7^4 + 9^4 + 10^4 = 1^4 + 2^4 + 5^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 9^4 + 10^4 = 2^4 + 2^4 + 2^4 + 3^4 + 5^4 + 8^4 + 11^4 = 2^4 + 4^4 + 4^4 + 5^4 + 6^4 + 7^4 + 11^4 = 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 8^4 + 10^4 = 3^4 + 4^4 + 5^4 + 6^4 + 6^4 + 6^4 + 11^4 = 3^4 + 5^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

			21252 is a term because 21252 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 1^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 7^4 + 8^4 + 9^4 + 9^4.

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

Cf. A345574, A345780, A345820, A345829, A345831, A345840, A346285.

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 == 8])
    for x in range(len(rets)):
        print(rets[x])

A344945 Numbers that are the sum of five fourth powers in exactly eight ways.

Original entry on oeis.org

534130, 654754, 663155, 729219, 737459, 742770, 758354, 810034, 813459, 816579, 831250, 906034, 930499, 954930, 1009954, 1055619, 1083955, 1099459, 1101859, 1103554, 1106019, 1157634, 1167794, 1180003, 1215394, 1217539, 1246354, 1253074, 1255539, 1278690

Offset: 1

David Consiglio, Jr., Jun 03 2021

nonn

Differs from A344944 at term 2 because 619090 = 1^4 + 2^4 + 18^4 + 22^4 + 23^4 = 1^4 + 3^4 + 4^4 + 8^4 + 28^4 = 1^4 + 11^4 + 14^4 + 22^4 + 24^4 = 2^4 + 2^4 + 8^4 + 17^4 + 27^4 = 2^4 + 13^4 + 13^4 + 18^4 + 26^4 = 3^4 + 6^4 + 12^4 + 16^4 + 27^4 = 4^4 + 12^4 + 14^4 + 23^4 + 23^4 = 9^4 + 12^4 + 16^4 + 21^4 + 24^4 = 14^4 + 16^4 + 18^4 + 19^4 + 23^4.

			534130 is a term because 534130 = 1^4 + 3^4 + 16^4 + 22^4 + 22^4  = 2^4 + 2^4 + 4^4 + 7^4 + 27^4  = 2^4 + 3^4 + 6^4 + 6^4 + 27^4  = 2^4 + 6^4 + 9^4 + 21^4 + 24^4  = 4^4 + 16^4 + 17^4 + 18^4 + 23^4  = 6^4 + 8^4 + 11^4 + 22^4 + 23^4  = 7^4 + 8^4 + 16^4 + 19^4 + 24^4  = 13^4 + 14^4 + 14^4 + 21^4 + 22^4.

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

Cf. A341892, A344925, A344943, A344944, A345184, A345820.

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, 5):
    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])

A345770 Numbers that are the sum of six cubes in exactly eight ways.

Original entry on oeis.org

1981, 2105, 2168, 2277, 2368, 2376, 2431, 2466, 2538, 2557, 2583, 2646, 2665, 2672, 2746, 2753, 2763, 2765, 2880, 2881, 2916, 2961, 2970, 2977, 2979, 2987, 3007, 3040, 3042, 3049, 3068, 3088, 3141, 3159, 3169, 3185, 3248, 3278, 3311, 3312, 3367, 3384, 3393

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345517 at term 8 because 2438 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 12^3 = 1^3 + 5^3 + 5^3 + 9^3 + 9^3 + 9^3 = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 2^3 + 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 2^3 + 4^3 + 5^3 + 8^3 + 9^3 + 10^3 = 4^3 + 4^3 + 5^3 + 5^3 + 9^3 + 11^3 = 6^3 + 7^3 + 7^3 + 8^3 + 8^3 + 8^3.

			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..1347

Cf. A345184, A345517, A345769, A345771, A345780, A345820.

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])

A346363 Numbers that are the sum of six fifth powers in exactly eight ways.

Original entry on oeis.org

2295937600, 4335900525, 6251954544, 8986552608, 13413708308, 14539246326, 15277569450, 15728636000, 16770321920, 16873011232, 17572402769, 17713454592, 17960776999, 18190647200, 19621666592, 20570070125, 20827689300, 22322555200, 23461554774, 23613244800

Offset: 1

David Consiglio, Jr., Jul 13 2021

nonn

This sequence differs from A345722:
9085584992 = 24^5 + 38^5 + 42^5 + 48^5 + 54^5 + 96^5
           = 21^5 + 34^5 + 38^5 + 43^5 + 74^5 + 92^5
           =  8^5 + 34^5 + 38^5 + 62^5 + 68^5 + 92^5
           = 18^5 + 18^5 + 44^5 + 64^5 + 66^5 + 92^5
           = 13^5 + 18^5 + 51^5 + 64^5 + 64^5 + 92^5
           =  8^5 + 38^5 + 41^5 + 47^5 + 79^5 + 89^5
           =  5^5 + 23^5 + 29^5 + 45^5 + 85^5 + 85^5
           =  8^5 + 23^5 + 41^5 + 64^5 + 82^5 + 84^5
           = 12^5 + 18^5 + 38^5 + 72^5 + 78^5 + 84^5,
so 9085584992 is in A345722, but is not in this sequence.

			2295937600 =  4^5 + 21^5 + 38^5 + 42^5 + 43^5 + 72^5
           =  8^5 + 16^5 + 30^5 + 42^5 + 54^5 + 70^5
           =  8^5 + 13^5 + 36^5 + 37^5 + 57^5 + 69^5
           = 14^5 + 16^5 + 16^5 + 52^5 + 54^5 + 68^5
           =  3^5 + 14^5 + 32^5 + 44^5 + 61^5 + 66^5
           =  4^5 + 18^5 + 22^5 + 52^5 + 58^5 + 66^5
           = 10^5 + 14^5 + 26^5 + 42^5 + 63^5 + 65^5
           =  1^5 +  7^5 + 34^5 + 57^5 + 58^5 + 63^5,
so 2295937600 is a term.

Sean A. Irvine, Table of n, a(n) for n = 1..4934

Cf. A345722, A345820, A346285, A346362, A346364.

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, 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])

cp's OEIS Frontend

A345565 Numbers that are the sum of six fourth powers in eight or more ways.

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A345819 Numbers that are the sum of six fourth powers in exactly seven ways.

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A345821 Numbers that are the sum of six fourth powers in exactly nine ways.

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A345830 Numbers that are the sum of seven fourth powers in exactly eight ways.

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A344945 Numbers that are the sum of five fourth powers in exactly eight ways.

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A345770 Numbers that are the sum of six cubes in exactly eight ways.

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A346363 Numbers that are the sum of six fifth powers in exactly eight ways.

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