%I #17 Jan 28 2014 07:01:12
%S 0,8,9,16,18,25,31,32,33,34,39,40,41,42,43,46,48,50,52,54,58,61,67,69,
%T 74,75,77,79,80,82,84,85,87,88,90,93,95,96,97,99,101,103,104,105,107,
%U 110,111,113,115,116,117,118,121,123,127,129
%N Numbers n with non-unique solution to n = +- 1^2 +- 2^2 +- ... +- k^2 with minimal k giving at least one solution.
%C The minimal k = A231015(n).
%C Complement of A231272.
%H Alois P. Heinz, <a href="/A231016/b231016.txt">Table of n, a(n) for n = 1..10000</a>
%H Andrica, D., Vacaretu, D., <a href="http://www.cs.ubbcluj.ro/~studia-m/2006-4/andrica.pdf">Representation theorems and almost unimodal sequences</a>, Studia Univ. Babes-Bolyai, Mathematica, Vol. LI, 4 (2006), 23-33.
%F { n : A231071(n) > 1 }.
%e 0 = 1 + 4 - 9 + 16 - 25 - 36 + 49 = sum with signs reversed, so 0 is a member.
%e 9 = - 1 - 4 + 9 + 16 + 25 - 36 = 1 + 4 + 9 - 16 - 25 + 36, so 9 is a member.
%e A000330(k) = k(k+1)(2k+1)/6 = 1^2 + 2^2 + ... + k^2 is not a member, for k > 0.
%p b:= proc(n, i) option remember; local m, t; m:= (1+(3+2*i)*i)*i/6;
%p if n>m then 0 elif n=m then 1 else
%p t:= b(abs(n-i^2), i-1);
%p if t>1 then return 2 fi;
%p t:= t+b(n+i^2, i-1); `if`(t>1, 2, t)
%p fi
%p end:
%p a:= proc(n) option remember; local m, k;
%p for m from 1+ `if`(n=1, -1, a(n-1)) do
%p for k while b(m, k)=0 do od;
%p if b(m, k)>1 then return m fi
%p od
%p end:
%p seq(a(n), n=1..80); # _Alois P. Heinz_, Nov 06 2013
%t b[n_, i_] := b[n, i] = Module[{m, t}, m = (1+(3+2*i)*i)*i/6; Which[n>m, 0, n == m, 1, True, t = b[Abs[n-i^2], i-1]; If[t>1, Return[2]]; t = t + b[n+i^2, i-1]; If[t>1, 2, t]]]; a[n_] := a[n] = Module[{m, k}, For[m = 1 + If[n == 1, -1, a[n-1]], True, m++, For[k = 1, b[m, k] == 0, k++]; If[b[m, k]>1, Return[m]]]]; Table[a[n], {n, 1, 80}] (* _Jean-François Alcover_, Jan 28 2014, after _Alois P. Heinz_ *)
%Y Cf. A000330, A231015, A231071, A231272.
%K nonn
%O 1,2
%A _Jonathan Sondow_, Nov 06 2013