A345187 -id:A345187 - cp's OEIS frontend

A345185 Numbers that are the sum of five third powers in nine or more ways.

5860, 6112, 6138, 6462, 6497, 6588, 6651, 6859, 6947, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8056, 8163, 8289, 8352, 8371, 8443, 8506, 8560, 8569, 8630, 8632, 8758, 8928, 8991, 9017, 9045, 9080, 9099

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

			6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3  = 1^3 + 3^3 + 7^3 + 12^3 + 14^3  = 1^3 + 6^3 + 6^3 + 7^3 + 16^3  = 2^3 + 2^3 + 9^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 11^3 + 15^3  = 2^3 + 8^3 + 9^3 + 9^3 + 14^3  = 3^3 + 3^3 + 3^3 + 4^3 + 17^3  = 3^3 + 5^3 + 8^3 + 11^3 + 14^3  = 8^3 + 8^3 + 8^3 + 11^3 + 12^3.

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

Cf. A341891, A344802, A345146, A345183, A345186, A345187, A345518.

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

A345155 Numbers that are the sum of four third powers in ten or more ways.

Original entry on oeis.org

21896, 36225, 46872, 48321, 48825, 51506, 52416, 53200, 55575, 58338, 58968, 59059, 60480, 62244, 66024, 67536, 67851, 70434, 70525, 71155, 72819, 73808, 76384, 76923, 77896, 78624, 78912, 81081, 81991, 85995, 87507, 88641, 90181, 90783, 91448, 91728, 92008

Offset: 1

David Consiglio, Jr., Jun 09 2021

nonn

			21896 is a term because 21896 = 1^3 + 11^3 + 19^3 + 22^3  = 2^3 + 2^3 + 12^3 + 26^3  = 2^3 + 3^3 + 19^3 + 23^3  = 2^3 + 5^3 + 15^3 + 25^3  = 3^3 + 10^3 + 16^3 + 24^3  = 3^3 + 17^3 + 19^3 + 19^3  = 4^3 + 6^3 + 20^3 + 22^3  = 5^3 + 8^3 + 14^3 + 25^3  = 7^3 + 11^3 + 17^3 + 23^3  = 8^3 + 9^3 + 19^3 + 22^3.

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

Cf. A025375, A344928, A345121, A345146, A345156, A345187.

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, 4):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 10])
for x in range(len(rets)):
    print(rets[x])

A345519 Numbers that are the sum of six cubes in ten or more ways.

Original entry on oeis.org

3104, 3215, 3222, 3267, 3313, 3339, 3374, 3430, 3465, 3491, 3493, 3528, 3547, 3584, 3645, 3654, 3698, 3708, 3736, 3745, 3752, 3754, 3761, 3771, 3780, 3789, 3817, 3824, 3843, 3862, 3869, 3878, 3880, 3888, 3906, 3915, 3923, 3943, 3950, 3969, 3995, 4004, 4014

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			3215 is a term because 3215 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 13^3 = 1^3 + 1^3 + 2^3 + 2^3 + 9^3 + 12^3 = 1^3 + 1^3 + 3^3 + 8^3 + 8^3 + 11^3 = 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 12^3 = 1^3 + 3^3 + 3^3 + 3^3 + 10^3 + 11^3 = 1^3 + 3^3 + 8^3 + 8^3 + 8^3 + 9^3 = 2^3 + 3^3 + 6^3 + 6^3 + 8^3 + 11^3 = 3^3 + 3^3 + 3^3 + 8^3 + 9^3 + 10^3 = 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 12^3 = 6^3 + 6^3 + 6^3 + 6^3 + 7^3 + 10^3.

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

Cf. A345187, A345477, A345506, A345518, A345567, A345772.

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 >= 10])
    for x in range(len(rets)):
        print(rets[x])

A341897 Numbers that are the sum of five fourth powers in ten or more ways.

Original entry on oeis.org

954979, 1205539, 1574850, 1713859, 1801459, 1863859, 1877394, 1882579, 2071939, 2109730, 2138419, 2142594, 2157874, 2225859, 2288179, 2419954, 2492434, 2495939, 2605314, 2663539, 2711394, 2784499, 2835939, 2847394, 2849859, 2880994, 2919154, 2924674, 3007474

Offset: 1

David Consiglio, Jr., Jun 04 2021

nonn

			954979 =  1^4 +  2^4 + 11^4 + 19^4 + 30^4
       =  1^4 +  7^4 + 18^4 + 25^4 + 26^4
       =  3^4 +  8^4 + 17^4 + 20^4 + 29^4
       =  4^4 +  8^4 + 13^4 + 25^4 + 27^4
       =  4^4 +  9^4 + 10^4 + 11^4 + 31^4
       =  6^4 +  6^4 + 15^4 + 21^4 + 29^4
       =  7^4 + 10^4 + 18^4 + 19^4 + 29^4
       = 11^4 + 11^4 + 20^4 + 22^4 + 27^4
       = 16^4 + 17^4 + 17^4 + 24^4 + 25^4
       = 18^4 + 19^4 + 20^4 + 23^4 + 23^4
so 954979 is a term.

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

Cf. A341891, A341898, A344928, A345187, A345567.

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 >= 10])
for x in range(len(rets)):
    print(rets[x])

A345188 Numbers that are the sum of five third powers in exactly ten ways.

Original entry on oeis.org

5860, 6588, 6651, 6859, 6947, 8056, 8289, 8569, 8758, 9045, 9099, 9227, 9414, 9612, 9829, 10009, 10277, 10485, 10522, 10529, 10800, 10963, 10970, 11008, 11061, 11089, 11241, 11385, 11458, 11656, 11719, 11782, 11817, 11845, 11934, 11990, 12016, 12060, 12088

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

Differs from A345187 at term 8 because 8371 = 1^3 + 1^3 + 11^3 + 11^3 + 16^3 = 1^3 + 4^3 + 5^3 + 12^3 + 17^3 = 1^3 + 8^3 + 9^3 + 11^3 + 16^3 = 3^3 + 3^3 + 4^3 + 15^3 + 15^3 = 3^3 + 3^3 + 8^3 + 8^3 + 18^3 = 3^3 + 3^3 + 3^3 + 5^3 + 19^3 = 3^3 + 7^3 + 9^3 + 9^3 + 17^3 = 4^3 + 6^3 + 6^3 + 11^3 + 17^3 = 5^3 + 9^3 + 10^3 + 11^3 + 15^3 = 6^3 + 6^3 + 12^3 + 13^3 + 13^3 = 8^3 + 8^3 + 9^3 + 9^3 + 16^3.

			6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3  = 1^3 + 4^3 + 6^3 + 13^3 + 14^3  = 1^3 + 5^3 + 8^3 + 8^3 + 16^3  = 1^3 + 10^3 + 10^3 + 11^3 + 12^3  = 2^3 + 2^3 + 9^3 + 12^3 + 14^3  = 2^3 + 3^3 + 8^3 + 11^3 + 15^3  = 3^3 + 8^3 + 8^3 + 11^3 + 14^3  = 3^3 + 3^3 + 5^3 + 10^3 + 16^3  = 5^3 + 5^3 + 8^3 + 10^3 + 15^3  = 8^3 + 9^3 + 10^3 + 10^3 + 12^3.

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

Cf. A294744, A341898, A345156, A345186, A345187, A345772.

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 == 10])
for x in range(len(rets)):
    print(rets[x])

A344803 Numbers that are the sum of five squares in ten or more ways.

Original entry on oeis.org

107, 109, 116, 125, 128, 131, 133, 134, 136, 139, 140, 142, 146, 147, 148, 149, 151, 152, 154, 155, 157, 158, 160, 163, 164, 166, 167, 168, 170, 171, 172, 173, 174, 175, 176, 178, 179, 181, 182, 183, 184, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196

Offset: 1

Sean A. Irvine, May 29 2021

nonn

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

Cf. A025375, A294744, A344802, A345187, A345477.

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A345185 Numbers that are the sum of five third powers in nine or more ways.

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A345155 Numbers that are the sum of four third powers in ten or more ways.

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A345519 Numbers that are the sum of six cubes in ten or more ways.

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A341897 Numbers that are the sum of five fourth powers in ten or more ways.

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A345188 Numbers that are the sum of five third powers in exactly ten ways.

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A344803 Numbers that are the sum of five squares in ten or more ways.

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