A345810 -id:A345810 - cp's OEIS frontend

A345556 Numbers that are the sum of ten cubes in eight or more ways.

623, 625, 630, 632, 644, 651, 658, 662, 665, 677, 684, 688, 695, 697, 699, 708, 714, 715, 721, 723, 725, 728, 730, 733, 734, 736, 740, 745, 747, 749, 751, 752, 754, 756, 757, 758, 759, 760, 764, 766, 769, 771, 773, 775, 776, 777, 778, 780, 782, 785, 786, 787

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			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..10000

Cf. A345547, A345555, A345557, A345601, A345810, A346807.

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])

A345800 Numbers that are the sum of nine cubes in exactly eight ways.

Original entry on oeis.org

744, 770, 805, 818, 840, 842, 844, 847, 866, 868, 877, 880, 883, 887, 894, 908, 909, 910, 911, 913, 915, 916, 920, 940, 945, 946, 948, 950, 952, 954, 955, 957, 961, 964, 965, 972, 976, 983, 987, 990, 1000, 1001, 1002, 1006, 1007, 1013, 1015, 1025, 1028, 1032

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345547 at term 9 because 859 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 6^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3 + 7^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 7^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 6^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 7^3.

Likely finite.

			770 is a term because 770 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.

Sean A. Irvine, Table of n, a(n) for n = 1..93

Cf. A345547, A345790, A345799, A345801, A345810, A345850.

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

A345860 Numbers that are the sum of ten fourth powers in exactly eight ways.

Original entry on oeis.org

6675, 6740, 6755, 6805, 6995, 7015, 7030, 7045, 7095, 7270, 7300, 7365, 7429, 7494, 7525, 7540, 7590, 7605, 7750, 7780, 7845, 7955, 8005, 8085, 8150, 8195, 8215, 8310, 8450, 8470, 8500, 8630, 8644, 8709, 8710, 8790, 8885, 8949, 9124, 9189, 9190, 9250, 9255

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345601 at term 5 because 6820 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.

			6740 is a term because 6740 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4.

Sean A. Irvine, Table of n, a(n) for n = 1..8900

Cf. A345601, A345810, A345850, A345859, A345861, A346353.

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

A345809 Numbers that are the sum of ten cubes in exactly seven ways.

Original entry on oeis.org

440, 473, 499, 506, 525, 532, 534, 567, 571, 584, 588, 597, 599, 604, 606, 627, 637, 639, 640, 656, 660, 663, 669, 670, 673, 680, 682, 689, 691, 693, 696, 701, 702, 704, 707, 717, 718, 719, 726, 729, 735, 738, 743, 744, 750, 755, 761, 762, 763, 770, 783, 784

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345555 at term 16 because 623 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 + 7^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3.

Likely finite.

			473 is a term because 473 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.

Sean A. Irvine, Table of n, a(n) for n = 1..78

Cf. A345555, A345799, A345808, A345810, A345859.

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

A345811 Numbers that are the sum of ten cubes in exactly nine ways.

Original entry on oeis.org

632, 651, 658, 688, 695, 714, 736, 740, 745, 752, 773, 778, 780, 790, 795, 799, 801, 812, 813, 815, 816, 818, 821, 823, 825, 841, 843, 849, 851, 852, 853, 855, 856, 857, 858, 864, 866, 873, 880, 882, 883, 885, 890, 891, 892, 899, 905, 908, 913, 922, 924, 926

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345557 at term 7 because 721 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 8^3 = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3.

Likely finite.

			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..79

Cf. A345557, A345801, A345810, A345812, A345861.

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])

cp's OEIS Frontend

A345556 Numbers that are the sum of ten cubes in eight or more ways.

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A345800 Numbers that are the sum of nine cubes in exactly eight ways.

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A345860 Numbers that are the sum of ten fourth powers in exactly eight ways.

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A345809 Numbers that are the sum of ten cubes in exactly seven ways.

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A345811 Numbers that are the sum of ten cubes in exactly nine ways.

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