A344945 -id:A344945 - cp's OEIS frontend

A345820 Numbers that are the sum of six fourth powers in exactly eight ways.

58035, 59780, 87746, 96195, 96450, 102371, 106451, 106515, 108035, 108275, 108290, 108771, 112370, 112931, 115251, 122835, 122850, 124691, 125971, 133395, 133571, 133586, 134675, 136931, 138275, 138595, 143650, 144755, 145826, 147491, 148820, 149571, 150115

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345565 at term 4 because 88595 = 1^4 + 4^4 + 5^4 + 12^4 + 13^4 + 14^4 = 1^4 + 6^4 + 6^4 + 11^4 + 12^4 + 15^4 = 1^4 + 7^4 + 8^4 + 9^4 + 10^4 + 16^4 = 2^4 + 8^4 + 9^4 + 9^4 + 12^4 + 15^4 = 2^4 + 10^4 + 11^4 + 11^4 + 12^4 + 13^4 = 4^4 + 6^4 + 6^4 + 9^4 + 13^4 + 15^4 = 5^4 + 6^4 + 7^4 + 8^4 + 11^4 + 16^4 = 7^4 + 7^4 + 10^4 + 11^4 + 12^4 + 14^4.

			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. A344945, A345565, A345770, A345819, A345821, A345830, A346363.

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])

A341892 Numbers that are the sum of five fourth powers in exactly nine ways.

Original entry on oeis.org

619090, 775714, 1100579, 1179379, 1186834, 1243699, 1357315, 1367539, 1373859, 1422595, 1431234, 1436419, 1511299, 1536019, 1699234, 1734899, 1839874, 1858594, 1880850, 1950355, 1951650, 1978915, 2044819, 2052899, 2069955, 2085139, 2101779, 2119459, 2133234

Offset: 1

David Consiglio, Jr., Jun 04 2021

nonn

Differs from A341781 at term 3 because
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.

			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
so 619090 is a term.

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

Cf. A341891, A341898, A344927, A344945, A345186, A345821.

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

A344925 Numbers that are the sum of four fourth powers in exactly eight ways.

Original entry on oeis.org

13155858, 26421474, 35965458, 39803778, 98926434, 128198994, 143776179, 156279618, 210493728, 237073554, 248075538, 255831858, 257931378, 269965938, 270289698, 292967619, 293579874, 295880274, 300120003, 301080243, 302115843, 305670834, 309742434, 331957458

Offset: 1

David Consiglio, Jr., Jun 02 2021

nonn

Differs from A344924 at term 24 because 328118259 = 2^4 + 77^4 + 109^4 + 111^4 = 8^4 + 79^4 + 93^4 + 121^4 = 18^4 + 79^4 + 97^4 + 119^4 = 21^4 + 77^4 + 98^4 + 119^4 = 27^4 + 77^4 + 94^4 + 121^4 = 34^4 + 77^4 + 89^4 + 123^4 = 46^4 + 57^4 + 103^4 + 119^4 = 49^4 + 77^4 + 77^4 + 126^4 = 61^4 + 66^4 + 77^4 + 127^4.

			13155858 is a term 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.

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

Cf. A344738, A344923, A344924, A344927, A344945, A345153.

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

A344943 Numbers that are the sum of five fourth powers in exactly seven ways.

Original entry on oeis.org

197779, 211059, 217154, 236675, 431155, 444019, 480739, 503539, 530659, 548994, 564979, 568450, 571539, 602450, 602770, 621859, 625635, 625939, 626194, 650659, 651954, 653059, 654130, 666739, 687314, 692754, 692899, 698019, 708499, 716739, 728914, 730914

Offset: 1

David Consiglio, Jr., Jun 03 2021

nonn

Differs from A344942 at term 10 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.

			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. A344923, A344941, A344942, A344945, A345181, A345819.

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])

A344944 Numbers that are the sum of five fourth powers in eight or more ways.

Original entry on oeis.org

534130, 619090, 654754, 663155, 729219, 737459, 742770, 758354, 775714, 810034, 813459, 816579, 831250, 906034, 930499, 954930, 954979, 1009954, 1055619, 1083955, 1099459, 1100579, 1101859, 1103554, 1106019, 1157634, 1167794, 1179379, 1180003, 1186834

Offset: 1

David Consiglio, Jr., Jun 03 2021

nonn

			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. A341891, A344922, A344924, A344942, A344945, A345183, A345565.

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])

A345184 Numbers that are the sum of five third powers in exactly eight ways.

Original entry on oeis.org

4392, 4915, 5139, 5256, 5321, 5624, 5643, 5678, 5741, 5769, 5797, 5832, 5914, 6075, 6202, 6499, 6560, 6616, 6642, 6677, 6833, 6884, 7008, 7111, 7128, 7155, 7218, 7344, 7395, 7641, 7696, 7729, 7785, 7813, 7820, 7849, 7883, 8037, 8100, 8243, 8282, 8308, 8315

Offset: 1

David Consiglio, Jr., Jun 10 2021

nonn

Differs from A345183 at term 13 because 5860 = 1^3 + 1^3 + 5^3 + 8^3 + 16^3 = 1^3 + 2^3 + 3^3 + 11^3 + 15^3 = 1^3 + 3^3 + 8^3 + 11^3 + 14^3 = 1^3 + 5^3 + 5^3 + 10^3 + 15^3 = 1^3 + 9^3 + 10^3 + 10^3 + 12^3 = 2^3 + 3^3 + 8^3 + 9^3 + 15^3 = 2^3 + 3^3 + 5^3 + 12^3 + 14^3 = 2^3 + 8^3 + 8^3 + 12^3 + 12^3 = 3^3 + 8^3 + 8^3 + 9^3 + 14^3 = 3^3 + 6^3 + 7^3 + 12^3 + 13^3.

			4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3  = 1^3 + 3^3 + 7^3 + 9^3 + 14^3  = 1^3 + 8^3 + 8^3 + 11^3 + 11^3  = 2^3 + 4^3 + 6^3 + 6^3 + 15^3  = 3^3 + 3^3 + 5^3 + 7^3 + 15^3  = 3^3 + 3^3 + 10^3 + 11^3 + 11^3  = 4^3 + 6^3 + 6^3 + 8^3 + 14^3  = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.

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

Cf. A294742, A344945, A345153, A345181, A345183, A345186, A345770.

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

cp's OEIS Frontend

A345820 Numbers that are the sum of six fourth powers in exactly eight ways.

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A341892 Numbers that are the sum of five fourth powers in exactly nine ways.

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A344925 Numbers that are the sum of four fourth powers in exactly eight ways.

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A344943 Numbers that are the sum of five fourth powers in exactly seven ways.

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A344944 Numbers that are the sum of five fourth powers in eight or more ways.

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A345184 Numbers that are the sum of five third powers in exactly eight ways.

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