A243113 Minimum of the cube root of the largest element over all partitions of n into at most 5 cubes.

0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 5, 3, 3, 3, 3, 3, 3, 4, 4, 5, 4, 3, 3, 3, 4, 4, 4, 4, 5, 4, 3, 4, 4, 3, 4, 4, 4, 4, 5

Offset: 0

Author

David S. Newman, Aug 20 2014

Keywords

• nonn

Comments

It is known that every number can be written as the sum of at most 5 (positive or negative) cubes.

"Minimum of the cube root of the largest absolute element over all partitions of n into at most 5 cubes" gives a different sequence with differences at n=302, 509, 517, 518, 521, 581, 733, 860, 1076, 1228, 1642, 1733, 1741, 1885, 2012, ... . - Alois P. Heinz, Aug 26 2014

Examples

			For n=5, a(n)=1. The partition of 5 into 1^3 + 1^3 + 1^3 + 1^3 + 1^3 has largest summand 1^3, while any other such partition, take 2^3 -1^3 -1^3 -1^3 for example, will have a larger largest part.
a(302) = 7: 7^3 +7^3 +4^3 +4^3 -8^3 = 302.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..20000

Programs

Maple

b:= proc(n, i, t) option remember; n=0 or (0<=i or n<=i^3)
      and t>0 and (b(n, i-1, t) or b(n-i^3, i, t-1))
    end:
a:= proc(n) local k; for k from 0
      do if b(n, k, 5) then return k fi od
    end:
seq(a(n), n=0..120);  # Alois P. Heinz, Aug 20 2014

Mathematica

b[n_, i_, t_] := b[n, i, t] = n==0 || (0 <= i || n <= i^3) && t>0 && (b[n, i-1, t] || b[n-i^3, i, t-1]); a[n_] := For[k=0, True, k++, If[b[n, k, 5], Return[k]]]; Table[a[n], {n, 0, 120}] (* Jean-François Alcover, Feb 17 2017, after Alois P. Heinz *)

Extensions

More terms from Alois P. Heinz, Aug 20 2014