A344943 -id:A344943 - cp's OEIS frontend

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

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])

A344923 Numbers that are the sum of four fourth powers in exactly seven ways.

Original entry on oeis.org

6576339, 16020018, 16408434, 22673634, 23056803, 33734834, 39786098, 43583138, 51071619, 52652754, 53731458, 57976083, 63985314, 64365939, 67655779, 68846274, 73744563, 75951138, 77495778, 87038883, 88648914, 89148114, 90665058, 90818898, 92800178, 93830803

Offset: 1

David Consiglio, Jr., Jun 02 2021

nonn

Differs from A344922 at term 2 because 13155858 = 1^4 + 16^4 + 19^4 + 60^4 = 3^4 + 6^4 + 21^4 + 60^4 = 10^4 + 18^4 + 31^4 + 59^4 = 12^4 + 27^4 + 45^4 + 54^4 = 15^4 + 44^4 + 46^4 + 47^4 = 18^4 + 25^4 + 41^4 + 56^4 = 29^4 + 30^4 + 44^4 + 53^4 = 35^4 + 36^4 + 38^4 + 53^4.

			6576339 is a term 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.

David Consiglio, Jr., Table of n, a(n) for n = 1..100

Cf. A344730, A344921, A344922, A344925, A344943, A345151.

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

A344941 Numbers that are the sum of five fourth powers in exactly six ways.

Original entry on oeis.org

151300, 225890, 236194, 243235, 246674, 250834, 286114, 288579, 300835, 302130, 302210, 303235, 309059, 317795, 320195, 334819, 334899, 335443, 336210, 338914, 346835, 356899, 363379, 366995, 373234, 375619, 389875, 391154, 392259, 393314, 394354, 412339

Offset: 1

David Consiglio, Jr., Jun 03 2021

nonn

Differs from A344940 at term 2 because 197779 = 1^4 + 5^4 + 6^4 + 16^4 + 19^4 = 1^4 + 7^4 + 11^4 + 12^4 + 20^4 = 1^4 + 10^4 + 12^4 + 17^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 21^4 = 3^4 + 5^4 + 6^4 + 6^4 + 21^4 = 4^4 + 7^4 + 9^4 + 13^4 + 20^4 = 11^4 + 13^4 + 14^4 + 15^4 + 16^4.

			151300 is a term because 151300 = 3^4 + 3^4 + 3^4 + 12^4 + 19^4  = 3^4 + 11^4 + 11^4 + 14^4 + 17^4  = 3^4 + 13^4 + 13^4 + 13^4 + 16^4  = 6^4 + 9^4 + 9^4 + 9^4 + 19^4  = 7^4 + 11^4 + 11^4 + 11^4 + 18^4  = 8^4 + 9^4 + 13^4 + 13^4 + 17^4.

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

Cf. A344359, A344921, A344940, A344943, A345175, A345818.

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

A344942 Numbers that are the sum of five fourth powers in seven or more ways.

Original entry on oeis.org

197779, 211059, 217154, 236675, 431155, 444019, 480739, 503539, 530659, 534130, 548994, 564979, 568450, 571539, 602450, 602770, 619090, 621859, 625635, 625939, 626194, 650659, 651954, 653059, 654130, 654754, 663155, 666739, 687314, 692754, 692899, 698019

Offset: 1

David Consiglio, Jr., Jun 03 2021

nonn

			197779 is a term because 197779 = 1^4 + 5^4 + 6^4 + 16^4 + 19^4  = 1^4 + 7^4 + 11^4 + 12^4 + 20^4  = 1^4 + 10^4 + 12^4 + 17^4 + 17^4  = 2^4 + 4^4 + 5^4 + 7^4 + 21^4  = 3^4 + 5^4 + 6^4 + 6^4 + 21^4  = 4^4 + 7^4 + 9^4 + 13^4 + 20^4  = 11^4 + 13^4 + 14^4 + 15^4 + 16^4.

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

Cf. A344922, A344940, A344943, A344944, A345180, A345564.

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 >= 7])
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])

A345181 Numbers that are the sum of five third powers in exactly seven ways.

Original entry on oeis.org

4472, 4544, 4600, 4957, 5076, 5113, 5120, 5132, 5165, 5174, 5347, 5354, 5384, 5391, 5410, 5445, 5474, 5481, 5507, 5543, 5617, 5715, 5760, 5834, 5895, 5923, 5984, 5986, 6049, 6128, 6131, 6245, 6280, 6373, 6407, 6434, 6436, 6544, 6553, 6733, 6768, 6831, 6840

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

Differs from A345180 at term 1 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.

			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. A294741, A344943, A345151, A345175, A345180, A345184, A345769.

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])

cp's OEIS Frontend

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

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A344923 Numbers that are the sum of four fourth powers in exactly seven ways.

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A344941 Numbers that are the sum of five fourth powers in exactly six ways.

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A344942 Numbers that are the sum of five fourth powers in seven or more ways.

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

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A345181 Numbers that are the sum of five third powers in exactly seven ways.

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