A025458 Number of partitions of n into 5 positive cubes.

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

Author

David W. Wilson

Keywords

• nonn

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

Links

• Robert Israel, Table of n, a(n) for n = 0..10000
• Index entries for sequences related to sums of cubes

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