A344244 -id:A344244 - cp's OEIS frontend

A344237 Numbers that are the sum of five fourth powers in exactly two ways.

260, 275, 340, 515, 884, 1555, 2595, 2660, 2675, 2690, 2705, 2755, 2770, 2835, 2930, 2945, 3010, 3185, 3299, 3314, 3379, 3554, 3923, 3970, 3985, 4050, 4115, 4145, 4160, 4210, 4290, 4355, 4400, 4465, 4594, 4769, 4834, 5075, 5090, 5155, 5265, 5330, 5395, 5440, 5505, 5570, 5699, 6370, 6545, 6580, 6595, 6660, 6675

Offset: 1

David Consiglio, Jr., May 12 2021

nonn

Differs from A344237 at term 31 because 4225 = 2^4 + 2^4 + 2^4 + 3^4 + 8^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4

			340 is a member of this sequence because 340 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4

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

Cf. A048927, A342686, A344190, A344193, A344238, A344244, A345814.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1,50)]
for pos in cwr(power_terms,5):
    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])

A344243 Numbers that are the sum of five fourth powers in three or more ways.

Original entry on oeis.org

4225, 6610, 6850, 9170, 9235, 9490, 11299, 12929, 14209, 14690, 14755, 14770, 15314, 16579, 16594, 16659, 16834, 17203, 17235, 17315, 17859, 17874, 17939, 18785, 18850, 18979, 19154, 19700, 19715, 20674, 20995, 21235, 21250, 21330, 21364, 21410, 21954, 23139, 23795, 24754, 25810, 26578, 28610, 28930

Offset: 1

David Consiglio, Jr., May 12 2021

nonn

			6850 = 1^4 + 2^4 + 2^4 + 4^4 + 9^4
     = 2^4 + 3^4 + 4^4 + 7^4 + 8^4
     = 3^4 + 3^4 + 6^4 + 6^4 + 8^4
so 6850 is a term of this sequence.

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

Cf. A342687, A343704, A344238, A344241, A344244, A344354, A345560.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1,50)]
for pos in cwr(power_terms,5):
    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])

A344355 Numbers that are the sum of five fourth powers in exactly four ways.

Original entry on oeis.org

20995, 21235, 31250, 41474, 43235, 43250, 43315, 43490, 43859, 45139, 46290, 47570, 51939, 53234, 53299, 54994, 56274, 57379, 57410, 57779, 59329, 63970, 67010, 68035, 68290, 71795, 71954, 73730, 73954, 75714, 75794, 77890, 82099, 84499, 86275, 86450, 87730, 92500, 93474, 93859, 94130, 94210, 96194

Offset: 1

David Consiglio, Jr., May 15 2021

nonn

Differs from A344354 at term 22 because 59779 = 1^4 + 1^4 + 5^4 + 12^4 + 14^4 = 1^4 + 6^4 + 6^4 + 9^4 + 15^4 = 2^4 + 9^4 + 10^4 + 11^4 + 13^4 = 4^4 + 7^4 + 7^4 + 8^4 + 15^4 = 7^4 + 7^4 + 9^4 + 10^4 + 14^4.

			31250 is a term of this sequence because 31250 = 2^4 + 2^4 + 4^4 + 7^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 13^4 = 4^4 + 6^4 + 7^4 + 9^4 + 12^4 = 5^4 + 5^4 + 10^4 + 10^4 + 10^4.

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

Cf. A344035, A344244, A344353, A344354, A344359, A344519, A345816.

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, 50)]
for pos in cwr(power_terms, 5):
    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])

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])

A342688 Numbers that are the sum of five positive fifth powers in exactly three ways.

Original entry on oeis.org

13124675, 28055699, 50043937, 52679923, 53069024, 55097976, 57936559, 60484744, 62260463, 62445305, 70211956, 73133026, 79401728, 80368962, 84766210, 88512249, 93288865, 98824300, 106993391, 113055482, 117173891, 120968132, 123383875, 126416258, 131106051, 131529588, 132022925

Offset: 1

David Consiglio, Jr., May 18 2021

nonn

Differs from A342687:
287618651 =  8^5 + 21^5 + 27^5 + 27^5 + 48^5
          =  9^5 + 13^5 + 26^5 + 37^5 + 46^5
          = 11^5 + 12^5 + 23^5 + 41^5 + 44^5
          = 11^5 + 20^5 + 22^5 + 30^5 + 48^5.
So 287618651 is a term of A342687 but not a term of this sequence.
[Corrected by Patrick De Geest, Dec 28 2024]

			50043937 =  6^5 + 16^5 + 18^5 + 24^5 + 33^5
         =  7^5 + 13^5 + 21^5 + 23^5 + 33^5
         = 11^5 + 13^5 + 13^5 + 29^5 + 31^5
so 50043937 is a term of this sequence.
[Corrected by _Patrick De Geest_, Dec 28 2024]

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

Cf. A003350, A004845, A342685, A342686, A342687, A344244, A344519, A344518, A345863, A345864, A346257, 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, 500)]
for pos in cwr(power_terms, 5):
    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])

A344242 Numbers that are the sum of four fourth powers in exactly three ways.

Original entry on oeis.org

16578, 43234, 49329, 53218, 54978, 57154, 93393, 106354, 107649, 108754, 138258, 151219, 160434, 168963, 173539, 177699, 178738, 181138, 183603, 185298, 195378, 195859, 196418, 197154, 197778, 201683, 202419, 209763, 211249, 216594, 217138, 223074, 234274, 235554, 235569, 237249, 237699, 240834

Offset: 1

David Consiglio, Jr., May 12 2021

nonn

Differs from A344241 at term 36 because 236674 = 1^4 + 2^4 + 7^4 + 22^4 = 3^4 + 6^4 + 18^4 + 19^4 = 7^4 + 14^4 + 16^4 + 19^4 = 8^4 + 16^4 + 17^4 + 17^4

			49329 is a member of this sequence because 49329 = 2^4 + 2^4 + 12^4 + 13^4 = 4^4 + 8^4 + 9^4 + 14^4 = 6^4 + 9^4 + 12^4 + 12^4

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

Cf. A025405, A344193, A344240, A344241, A344244, A344353.

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

A344519 Numbers that are the sum of five positive fifth powers in exactly four ways.

Original entry on oeis.org

287618651, 1386406515, 1763135232, 2494769760, 2619898293, 3096064443, 3291315732, 3749564512, 4045994624, 5142310350, 5183605813, 5658934676, 5880926107, 7205217018, 7401155424, 7691215599, 8429499101, 8926086432, 9051501568, 9203796832, 9254212901

Offset: 1

David Consiglio, Jr., May 21 2021

nonn

Differs from A344518 at term 20 because
9006349824 =  8^5 + 34^5 + 62^5 + 68^5 + 92^5
           =  8^5 + 41^5 + 47^5 + 79^5 + 89^5
           = 12^5 + 18^5 + 72^5 + 78^5 + 84^5
           = 21^5 + 34^5 + 43^5 + 74^5 + 92^5
           = 24^5 + 42^5 + 48^5 + 54^5 + 96^5.

			287618651 is a term because
287618651 =  8^5 + 21^5 + 27^5 + 27^5 + 48^5
          =  9^5 + 13^5 + 26^5 + 37^5 + 46^5
          = 11^5 + 12^5 + 23^5 + 41^5 + 44^5
          = 11^5 + 20^5 + 22^5 + 30^5 + 48^5.
[Corrected by _Patrick De Geest_, Dec 28 2024]

Sean A. Irvine, Table of n, a(n) for n = 1..4857

Cf. A003350, A004845, A342685, A342686, A342687, A342688, A344244, A344355, A344518, A345863, A345864, A346257, A346358, A346359.

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, 500)]
for pos in cwr(power_terms, 5):
    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])

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A344237 Numbers that are the sum of five fourth powers in exactly two ways.

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A344243 Numbers that are the sum of five fourth powers in three or more ways.

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A344355 Numbers that are the sum of five fourth powers in exactly four ways.

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

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A342688 Numbers that are the sum of five positive fifth powers in exactly three ways.

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A344242 Numbers that are the sum of four fourth powers in exactly three ways.

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A344519 Numbers that are the sum of five positive fifth powers in exactly four ways.

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