login
Cube root of the smallest of the largest absolute values of parts of the partitions of n into four cubes, or -1 if no such partition exists.
2

%I #19 Nov 13 2020 02:32:46

%S 0,1,1,1,1,2,2,2,2,2,2,2,3,11,2,2,2,2,2,3,3,3,16,2,2,2,3,3,3,3,3,52,2,

%T 3,3,3,3,3,3,4,4,8,3,3,3,3,3,3,4,4,49,3,3,3,3,3,3,4,4,5,5,3,3,3,4,4,4,

%U 4,5,5,3,4,4,3,4,4,11,5,8,4,3,3,3,4,4

%N Cube root of the smallest of the largest absolute values of parts of the partitions of n into four cubes, or -1 if no such partition exists.

%C It is not known if every integer can be written as the sum of four cubes, but it is true at least up to 1000 by computer search.

%C For each partition of n into four cubes (positive, negative, or zero) choose the largest part in absolute value. a(n) is the cube root of the smallest such largest part over all such partitions.

%C If there is no partition of n into four cubes, then a(n) = -1.

%C There is an interesting correlation with A332201 (sum of three cubes problem) whose nonzero absolute values coincide with a(n+1) up to n=30. - _M. F. Hasler_, Feb 10 2020

%H Alois P. Heinz, <a href="/A246869/b246869.txt">Table of n, a(n) for n = 0..20000</a>

%e The partition of 13 into 1^3+7^3+10^3+(-11)^3 has a part 11^3 in absolute value. Any other partition of 13 into four cubes has a part larger than 11^3 in absolute value. Thus a(13) = 11.

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

%p or b(n+i^3, i, t-1) or b(abs(n-i^3), i, t-1))

%p end:

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

%p do if b(n, k, 4) then return k fi od

%p end:

%p seq(a(n), n=0..30); # _Alois P. Heinz_, Sep 05 2014

%t b[n_, i_, t_] := b[n, i, t] = n == 0 || t i^3 >= n && (b[n, i - 1, t] || b[n + i^3, i, t - 1] || b[Abs[n - i^3], i, t - 1]);

%t a[n_] := Module[{k}, For[k = 0, True, k++, If[b[n, k, 4], Return[k]]]];

%t a /@ Range[0, 100] (* _Jean-François Alcover_, Nov 13 2020, after _Alois P. Heinz_ *)

%Y Cf. A243113.

%K nonn

%O 0,6

%A _David S. Newman_, Sep 05 2014

%E More terms from _Alois P. Heinz_, Sep 05 2014