A345811 -id:A345811 - cp's OEIS frontend

A345801 Numbers that are the sum of nine cubes in exactly nine ways.

859, 861, 896, 903, 922, 929, 935, 939, 959, 973, 997, 999, 1009, 1016, 1020, 1023, 1027, 1029, 1030, 1034, 1035, 1036, 1037, 1041, 1046, 1059, 1060, 1064, 1065, 1066, 1067, 1071, 1072, 1079, 1086, 1091, 1105, 1116, 1131, 1138, 1146, 1152, 1155, 1157, 1158

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345548 at term 10 because 966 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 6^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 8^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 9^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 + 7^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 + 7^3 = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 9^3 = 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3 = 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3.

Likely finite.

			861 is a term because 861 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 8^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3.

Sean A. Irvine, Table of n, a(n) for n = 1..86

Cf. A345548, A345791, A345800, A345802, A345811, A345851.

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, 9):
    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])

A345861 Numbers that are the sum of ten fourth powers in exactly nine ways.

Original entry on oeis.org

6820, 6870, 6950, 7060, 7110, 7125, 7285, 7350, 7860, 7925, 8020, 8230, 8245, 8260, 8325, 8390, 8405, 8645, 8755, 8820, 8884, 8965, 8995, 9030, 9045, 9060, 9075, 9125, 9220, 9270, 9365, 9430, 9475, 9490, 9525, 9605, 9730, 9735, 9765, 9815, 9895, 9910, 10035

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345602 at term 3 because 6885 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 9^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

			6870 is a term because 6870 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 9^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 2^4 + 2^4 + 2^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 6^4 + 7^4.

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

Cf. A345602, A345811, A345851, A345860, A345862, A346354.

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, 10):
    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])

A345557 Numbers that are the sum of ten cubes in nine or more ways.

Original entry on oeis.org

632, 651, 658, 688, 695, 714, 721, 736, 740, 745, 752, 754, 756, 771, 773, 778, 780, 782, 790, 792, 795, 797, 799, 801, 806, 808, 812, 813, 815, 816, 818, 819, 821, 823, 825, 832, 834, 841, 843, 845, 847, 848, 849, 850, 851, 852, 853, 855, 856, 857, 858, 860

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			651 is a term because 651 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3.

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

Cf. A345548, A345556, A345558, A345602, A345811.

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, 10):
    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])

A345810 Numbers that are the sum of ten cubes in exactly eight ways.

Original entry on oeis.org

623, 625, 630, 644, 662, 665, 677, 684, 697, 699, 708, 715, 723, 725, 728, 730, 733, 734, 747, 749, 751, 757, 758, 759, 760, 764, 766, 769, 775, 776, 777, 785, 786, 787, 789, 793, 794, 796, 804, 810, 811, 814, 817, 820, 826, 827, 828, 829, 830, 831, 836, 838

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345556 at term 4 because 632 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3.

Likely finite.

			625 is a term because 625 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 5^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.

Sean A. Irvine, Table of n, a(n) for n = 1..75

Cf. A345556, A345800, A345809, A345811, A345860.

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, 10):
    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])

A345812 Numbers that are the sum of ten cubes in exactly ten ways.

Original entry on oeis.org

721, 754, 756, 782, 792, 797, 806, 808, 819, 834, 847, 848, 850, 860, 871, 874, 876, 877, 878, 881, 884, 886, 893, 902, 903, 907, 909, 910, 916, 917, 918, 921, 929, 932, 933, 936, 937, 938, 941, 942, 944, 945, 955, 965, 966, 968, 973, 991, 994, 999, 1001

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345558 at term 4 because 771 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 9^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 4^3 + 7^3 + 7^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 7^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3.

Likely finite.

			754 is a term because 754 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3.

Sean A. Irvine, Table of n, a(n) for n = 1..72

Cf. A345558, A345802, A345811, A345862.

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, 10):
    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])

cp's OEIS Frontend

A345801 Numbers that are the sum of nine cubes in exactly nine ways.

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A345861 Numbers that are the sum of ten fourth powers in exactly nine ways.

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A345557 Numbers that are the sum of ten cubes in nine or more ways.

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A345810 Numbers that are the sum of ten cubes in exactly eight ways.

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A345812 Numbers that are the sum of ten cubes in exactly ten ways.

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