A243113 - OEIS

%I #29 Feb 17 2017 02:34:05

%S 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,

%T 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,

%U 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

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

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

%C "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

%H Alois P. Heinz, <a href="/A243113/b243113.txt">Table of n, a(n) for n = 0..20000</a>

%e 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.

%e a(302) = 7: 7^3 +7^3 +4^3 +4^3 -8^3 = 302.

%p b:= proc(n, i, t) option remember; n=0 or (0<=i or n<=i^3)

%p       and t>0 and (b(n, i-1, t) or b(n-i^3, i, t-1))

%p     end:

%p a:= proc(n) local k; for k from 0

%p       do if b(n, k, 5) then return k fi od

%p     end:

%p seq(a(n), n=0..120);  # _Alois P. Heinz_, Aug 20 2014

%t 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_ *)

%K nonn

%O 0,7

%A _David S. Newman_, Aug 20 2014

%E More terms from _Alois P. Heinz_, Aug 20 2014