A345149 -id:A345149 - cp's OEIS frontend

A345148 Numbers that are the sum of four third powers in six or more ways.

6883, 12411, 13104, 13923, 14112, 14581, 14896, 14904, 15561, 15876, 16317, 16640, 17208, 17479, 17992, 18739, 18865, 18928, 19035, 19080, 19376, 19665, 19712, 19763, 19880, 20007, 20384, 20755, 20979, 21203, 21231, 21420, 21707, 21896, 22409, 22617, 22743

Offset: 1

David Consiglio, Jr., Jun 09 2021

nonn

			6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3  = 2^3 + 4^3 + 14^3 + 14^3  = 3^3 + 7^3 + 7^3 + 17^3  = 3^3 + 10^3 + 13^3 + 13^3  = 4^3 + 10^3 + 10^3 + 15^3  = 7^3 + 8^3 + 8^3 + 16^3.

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

Cf. A025371, A343987, A344904, A345083, A345149, A345150, A345174.

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 >= 6])
for x in range(len(rets)):
    print(rets[x])

A343986 Numbers that are the sum of four positive cubes in exactly five ways.

Original entry on oeis.org

5105, 5131, 5616, 5859, 6435, 7777, 9315, 9737, 9793, 10017, 10250, 10458, 10936, 10962, 11000, 11060, 11088, 11592, 11664, 11781, 12168, 12229, 12285, 12320, 12385, 12392, 12707, 13384, 13734, 13832, 13904, 14183, 14239, 14833, 15176, 15596, 15624, 15752, 15759, 15778, 16093, 16289, 16354, 16480, 16569

Offset: 1

David Consiglio, Jr., May 06 2021

nonn

Differs from A343987 at term 6 because 6883 = 2^3 + 2^3 + 2^3 + 19^3 = 2^3 + 5^3 + 15^3 + 15^3 = 3^3 + 8^3 + 8^3 + 18^3 = 4^3 + 11^3 + 14^3 + 14^3 = 5^3 + 11^3 + 11^3 + 16^3 = 8^3 + 9^3 + 9^3 + 17^3.

			5616 is a term because 5616 = 1^3 + 8^3 + 12^3 + 15^3 = 2^3 + 8^3 + 10^3 + 16^3 = 4^3 + 4^3 + 14^3 + 14^3 = 4^3 + 5^3 + 11^3 + 16^3 = 8^3 + 9^3 + 10^3 + 15^3.

David Consiglio, Jr., Table of n, a(n) for n = 1..20000

Cf. A025361, A343970, A343972, A343987, A343988, A344357, A345149.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1,50)]
for pos in cwr(power_terms,4):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k,v in keep.items() if v == 5])
for x in range(len(rets)):
    print(rets[x])

A344921 Numbers that are the sum of four fourth powers in exactly six ways.

Original entry on oeis.org

3847554, 5624739, 6044418, 6593538, 6899603, 9851058, 10456338, 11645394, 12378018, 13638738, 16990803, 19081089, 20622338, 20649603, 20755218, 20795763, 24174003, 24368769, 25265553, 25850178, 25899058, 28470339, 29195154, 30295539, 30534018, 30623394

Offset: 1

David Consiglio, Jr., Jun 02 2021

nonn

Differs from A344904 at term 4 because 6576339 = 1^4 + 24^4 + 41^4 + 43^4 = 3^4 + 7^4 + 41^4 + 44^4 = 4^4 + 23^4 + 27^4 + 49^4 = 6^4 + 31^4 + 41^4 + 41^4 = 7^4 + 11^4 + 36^4 + 47^4 = 7^4 + 21^4 + 28^4 + 49^4 = 12^4 + 17^4 + 29^4 + 49^4.

			3847554 is a term because 3847554 = 2^4 + 13^4 + 29^4 + 42^4  = 2^4 + 21^4 + 22^4 + 43^4  = 6^4 + 11^4 + 17^4 + 44^4  = 6^4 + 31^4 + 32^4 + 37^4  = 9^4 + 29^4 + 32^4 + 38^4  = 13^4 + 26^4 + 32^4 + 39^4.

David Consiglio, Jr., Table of n, a(n) for n = 1..1000

Cf. A344357, A344648, A344904, A344923, A344941, A345149.

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, 4):
    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])

A345151 Numbers that are the sum of four third powers in exactly seven ways.

Original entry on oeis.org

13104, 18928, 19376, 20755, 21203, 22743, 24544, 24570, 24787, 25172, 25928, 27755, 27846, 28917, 29582, 31031, 31248, 31528, 32858, 34056, 34713, 35289, 35317, 35441, 35497, 35712, 36190, 36288, 36610, 36890, 36946, 38080, 39221, 39440, 39464, 39851, 39942

Offset: 1

David Consiglio, Jr., Jun 09 2021

nonn

Differs from A345150 at term 6 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.

			13104 is a term because 13104 = 1^3 + 10^3 + 16^3 + 18^3  = 1^3 + 11^3 + 14^3 + 19^3  = 2^3 + 9^3 + 15^3 + 19^3  = 4^3 + 6^3 + 14^3 + 20^3  = 4^3 + 9^3 + 10^3 + 21^3  = 5^3 + 7^3 + 11^3 + 21^3  = 8^3 + 9^3 + 14^3 + 19^3.

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

Cf. A025363, A344923, A345085, A345149, A345150, A345153, A345181.

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

A345175 Numbers that are the sum of five third powers in exactly six ways.

Original entry on oeis.org

2430, 2979, 3214, 3249, 3312, 3492, 3520, 3737, 3753, 3788, 3816, 3842, 3942, 3968, 4121, 4185, 4213, 4267, 4355, 4411, 4418, 4446, 4453, 4456, 4465, 4482, 4509, 4563, 4626, 4663, 4670, 4723, 4753, 4896, 4905, 4924, 4938, 4941, 4950, 4960, 4976, 4987, 4994

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

Differs from A345174 at term 20 because 4392 = 1^3 + 1^3 + 10^3 + 10^3 + 11^3 = 1^3 + 2^3 + 2^3 + 9^3 + 14^3 = 1^3 + 8^3 + 9^3 + 10^3 + 10^3 = 2^3 + 2^3 + 3^3 + 5^3 + 15^3 = 2^3 + 3^3 + 5^3 + 8^3 + 14^3 = 2^3 + 8^3 + 8^3 + 8^3 + 12^3 = 3^3 + 6^3 + 7^3 + 8^3 + 13^3 = 5^3 + 5^3 + 5^3 + 9^3 + 13^3.

			2430 is a term because 2430 = 1^3 + 2^3 + 2^3 + 5^3 + 12^3  = 1^3 + 3^3 + 4^3 + 7^3 + 11^3  = 2^3 + 2^3 + 6^3 + 6^3 + 11^3  = 2^3 + 3^3 + 3^3 + 9^3 + 10^3  = 3^3 + 5^3 + 8^3 + 8^3 + 8^3  = 3^3 + 4^3 + 7^3 + 8^3 + 9^3.

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

Cf. A294740, A343988, A344941, A345149, A345174, A345181, 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, 5):
    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])

A345084 Numbers that are the sum of three third powers in exactly six ways.

Original entry on oeis.org

1296378, 1371735, 1409400, 1614185, 1824040, 1885248, 2101464, 2302028, 2305395, 2542968, 2851848, 2889216, 2974392, 2988441, 3185792, 3380833, 3681280, 3689496, 3706984, 3775680, 3906657, 4109832, 4123008, 4142683, 4422592, 4842872, 4952312, 5005125, 5023656

Offset: 1

David Consiglio, Jr., Jun 07 2021

nonn

Differs from A345083 at term 7 because 2016496 = 5^3 + 71^3 + 117^3 = 9^3 + 65^3 + 119^3 = 18^3 + 20^3 + 125^3 = 46^3 + 96^3 + 99^3 = 53^3 + 59^3 + 117^3 = 65^3 + 89^3 + 99^3 = 82^3 + 84^3 + 93^3.

			1296378 is a term because 1296378 = 3^3 + 75^3 + 94^3  = 8^3 + 32^3 + 107^3  = 20^3 + 76^3 + 93^3  = 30^3 + 58^3 + 101^3  = 32^3 + 80^3 + 89^3  = 59^3 + 74^3 + 86^3.

David Consiglio, Jr., Table of n, a(n) for n = 1..1000

Cf. A025326, A343970, A344648, A345083, A345085, A345149.

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, 3):
    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])

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A345148 Numbers that are the sum of four third powers in six or more ways.

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A343986 Numbers that are the sum of four positive cubes in exactly five ways.

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A344921 Numbers that are the sum of four fourth powers in exactly six ways.

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A345151 Numbers that are the sum of four third powers in exactly seven ways.

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A345175 Numbers that are the sum of five third powers in exactly six ways.

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A345084 Numbers that are the sum of three third powers in exactly six ways.

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