cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A344034 Numbers that are the sum of five positive cubes in four or more ways.

Original entry on oeis.org

1252, 1376, 1461, 1522, 1548, 1585, 1590, 1646, 1702, 1709, 1737, 1739, 1765, 1772, 1798, 1802, 1810, 1864, 1889, 1954, 1980, 1987, 2006, 2033, 2043, 2081, 2096, 2104, 2152, 2160, 2195, 2225, 2241, 2250, 2251, 2276, 2313, 2322, 2339, 2341, 2367, 2374, 2377, 2416, 2423, 2430, 2449, 2456, 2458, 2465, 2467, 2486
Offset: 1

Views

Author

David Consiglio, Jr., May 07 2021

Keywords

Examples

			1461 = 1^3 + 1^3 + 1^3 + 9^3 +  9^3
     = 1^3 + 1^3 + 4^3 + 4^3 + 11^3
     = 3^3 + 3^3 + 4^3 + 7^3 + 10^3
     = 6^3 + 6^3 + 7^3 + 7^3 +  7^3
so 1461 is a term of this sequence.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**3 for x in range(1,50)]
    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])

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

Views

Author

David Consiglio, Jr., May 15 2021

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

  • Python
    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])

A343972 Numbers that are the sum of four positive cubes in exactly four ways.

Original entry on oeis.org

1979, 2737, 3663, 4384, 4445, 4474, 4949, 5257, 5320, 5473, 5499, 5553, 5733, 5768, 5833, 5852, 6064, 6104, 6328, 6372, 6587, 6643, 6832, 6912, 6974, 7000, 7030, 7120, 7217, 7371, 7560, 7686, 7840, 8099, 8108, 8281, 8316, 8344, 8379, 8414, 8505, 8568, 8927, 9016, 9018, 9044, 9072, 9100, 9289, 9548, 9648, 9800
Offset: 1

Views

Author

David Consiglio, Jr., May 05 2021

Keywords

Comments

This sequence varies from A343971 at term 8 because 5105 = 1^3 + 1^3 + 12^3 + 15^3 = 1^3 + 2^3 + 10^3 + 16^3 = 1^3 + 9^3 + 10^3 + 15^3 = 4^3 + 4^3 + 4^3 + 17^3 = 4^3 + 6^3 + 9^3 + 16^3.

Examples

			3663 is a term because 3663 = 1^3 + 10^3 + 11^3 + 11^3 = 2^3 + 4^3 + 6^3 + 15^3 = 2^3 + 9^3 + 9^3 + 13^3 = 4^3 + 7^3 + 8^3 + 14^3.
		

Crossrefs

Programs

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

A343988 Numbers that are the sum of five positive cubes in exactly five ways.

Original entry on oeis.org

1765, 1980, 2043, 2104, 2195, 2250, 2449, 2486, 2491, 2493, 2547, 2584, 2592, 2738, 2745, 2764, 2817, 2888, 2915, 2953, 2969, 3095, 3096, 3133, 3142, 3186, 3188, 3240, 3275, 3277, 3310, 3366, 3403, 3422, 3459, 3464, 3466, 3483, 3529, 3583, 3608, 3627, 3653, 3664, 3671, 3690, 3697, 3707, 3725, 3744, 3746, 3781
Offset: 1

Views

Author

David Consiglio, Jr., May 06 2021

Keywords

Comments

Differs from A343989 at term 7 because 2430 = 1^3 + 2^3 + 2^3 + 6^3 + 13^3 = 1^3 + 4^3 + 5^3 + 8^3 + 12^3 = 2^3 + 2^3 + 7^3 + 7^3 + 12^3 = 2^3 + 3^3 + 4^3 + 10^3 + 11^3 = 3^3 + 6^3 + 9^3 + 9^3 + 9^3 = 4^3 + 5^3 + 8^3 + 9^3 + 10^3.

Examples

			2043 is a term because 2043 = 1^3 + 4^3 + 5^3 + 5^3 + 12^3 = 2^3 + 2^3 + 3^3 + 10^3 + 10^3 = 2^3 + 3^3 + 4^3 + 6^3 + 12^3 = 4^3 + 5^3 + 5^3 + 9^3 + 10^3 = 4^3 + 6^3 + 6^3 + 6^3 + 11^3.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**3 for x in range(1,50)]
    for pos in cwr(power_terms,5):
        tot = sum(pos)
        keep[tot] += 1
    rets = sorted([k for k,v in keep.items() if v == 5])
    for x in range(len(rets)):
        print(rets[x])

A345766 Numbers that are the sum of six cubes in exactly four ways.

Original entry on oeis.org

626, 830, 837, 856, 873, 891, 947, 954, 982, 1008, 1026, 1052, 1053, 1071, 1094, 1097, 1106, 1109, 1134, 1143, 1150, 1153, 1172, 1195, 1208, 1227, 1234, 1253, 1267, 1278, 1279, 1283, 1286, 1290, 1297, 1316, 1323, 1324, 1358, 1361, 1368, 1369, 1376, 1395, 1403
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345513 at term 12 because 1045 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 10^3 = 1^3 + 1^3 + 4^3 + 5^3 + 5^3 + 9^3 = 1^3 + 2^3 + 3^3 + 4^3 + 6^3 + 9^3 = 3^3 + 3^3 + 6^3 + 6^3 + 6^3 + 7^3 = 4^3 + 4^3 + 4^3 + 5^3 + 6^3 + 8^3.

Examples

			830 is a term because 830 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 8^3 = 1^3 + 3^3 + 3^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 6^3 + 6^3.
		

Crossrefs

Programs

  • Python
    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, 6):
        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])

A025458 Number of partitions of n into 5 positive cubes.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Keywords

Comments

a(n) > 2 at n= 766, 810, 827, 829, 865, 883, 981, 1018, 1025, 1044,... - R. J. Mathar, Sep 15 2015
The first term > 1 is a(157) = 2. - Michel Marcus, Apr 25 2019

Crossrefs

Column 5 of A320841, which cross-references the equivalent sequences for other numbers of positive cubes.
Positions of values: A057906 (0), A003328 (nonzero), A048926 (1), A048927 (2), A343705 (3), A344035 (4).

Programs

  • Maple
    A025458 := proc(n)
        local a,x,y,z,u,vcu ;
        a := 0 ;
        for x from 1 do
            if 5*x^3 > n then
                return a;
            end if;
            for y from x do
                if x^3+4*y^3 > n then
                    break;
                end if;
                for z from y do
                    if x^3+y^3+3*z^3 > n then
                        break;
                    end if;
                    for u from z do
                        if x^3+y^3+z^3+2*u^3 > n then
                            break;
                        end if;
                        vcu := n-x^3-y^3-z^3-u^3 ;
                        if isA000578(vcu) then
                            a := a+1 ;
                        end if;
                    end do:
                end do:
            end do:
        end do:
    end proc: # R. J. Mathar, Sep 15 2015
  • Mathematica
    a[n_] := IntegerPartitions[n, {5}, Range[n^(1/3) // Ceiling]^3] // Length;
    a /@ Range[0, 157] (* Jean-François Alcover, Jun 20 2020 *)

Formula

a(n) = [x^n y^5] Product_{k>=1} 1/(1 - y*x^(k^3)). - Ilya Gutkovskiy, Apr 23 2019

Extensions

Second offset from Michel Marcus, Apr 25 2019
Showing 1-6 of 6 results.