A025469 Number of partitions of n into 3 distinct positive cubes.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Author

David W. Wilson

Keywords

• nonn

Comments

In other words, number of solutions to the equation n = x^3 + y^3 + z^3 with x > y > z > 0. - Antti Karttunen, Aug 29 2017

Examples

			From _Antti Karttunen_, Aug 29 2017: (Start)
For n = 36 there is one solution: 36 = 27 + 8 + 1, thus a(36) = 1.
For n = 1009 there are two solutions: 1009 = 10^3 + 2^3 + 1^3 = 9^3 + 6^3 + 4^3, thus a(1009) = 2. This is also the first point where sequence attains value greater than one.
(End)
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..17073
• Index entries for sequences related to sums of cubes

Crossrefs

Cf. A025465 (not necessarily distinct), A025468, A025419 (greedy inverse).

Cf. A024975 (positions of nonzero terms), A024974 (positions of terms > 1), A025399-A025402.

Programs

Maple

A025469 := proc(n)
    local a, x, y, zcu ;
    a := 0 ;
    for x from 1 do
        if 3*x^3 > n then
            return a;
        end if;
        for y from x+1 do
            if x^3+2*y^3 > n then
                break;
            end if;
            zcu := n-x^3-y^3 ;
            if zcu > y^3 and isA000578(zcu) then
                a := a+1 ;
            end if;
        end do:
    end do:
end proc:
seq(A025469(n),n=1..80) ; # R. J. Mathar, Jun 15 2018

Mathematica

Table[Count[IntegerPartitions[n, {3}], ?(And[UnsameQ @@ #, AllTrue[#, IntegerQ[#^(1/3)] &]] &)], {n, 105}] (* _Michael De Vlieger, Aug 29 2017 *)

PARI

A025469(n) = { my(s=0); for(x=1,n,if(ispower(x,3),for(y=x+1,n-x,if(ispower(y,3),for(z=y+1,n-(x+y),if((ispower(z,3)&&(x+y+z)==n),s++)))))); (s); }; \\ Antti Karttunen, Aug 29 2017

Formula

a(n) = A025465(n) - A025468(n). - Antti Karttunen, Aug 29 2017