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%I #27 Nov 12 2020 06:32:08
%S 1,2,3,2,5,4,3,4,3,4,5,6,5,4,5,4,5,6,5,6,7,6,5,6,5,6,7,6,7,6,7,8,7,6,
%T 7,6,7,8,7,8,7,8,7,8,9,8,7,8,7,8,9,8,9,8,9,8,9,8,9,10,9,8,9,8,9,10,9,
%U 10,9,10,9,10,9,10,9,10,11,10,9,10,9,10,11,10,11,10,11,10,11,10,11,10,11,10,11,12,11,10,11,10
%N Smallest positive integer k such that n = +-1 +-3 +-... +-(2k-1) for some choice of +'s and -'s
%C Conjecture. Let SO(k) be the sum of the first k odd positive integers. Then a(n)=k if n=SO(k). Otherwise, choose k so that SO(k-1)<n<SO(k). Then if SO(k)-n=4, a(n)=k+2, else if SO(k)-n is odd then a(n)=k+1 else a(n)=k. (This has been verified for n up to 200.)
%H Alois P. Heinz, <a href="/A197702/b197702.txt">Table of n, a(n) for n = 1..10000</a>
%e The sum of 3 terms 1 - 3 + 5 gives 3, but none of the 2-term sums 1+3, 1-3, -1+3, -1-3 gives 3, so a(3)=3.
%p b:= proc(n, i) option remember; (n=0 and i=0) or
%p abs(n)<=i^2 and (b(n-2*i+1, i-1) or b(n+2*i-1, i-1))
%p end:
%p a:= proc(n) local k;
%p for k from floor(sqrt(n)) while not b(n, k) do od; k
%p end:
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Oct 19 2011
%t b[n_, i_] := b[n, i] = (n==0 && i==0) || Abs[n] <= i^2 && (b[n-2i+1, i-1] || b[n+2i-1, i-1]);
%t a[n_] := Module[{k}, For[k = Floor[Sqrt[n]], !b[n, k], k++]; k];
%t Array[a, 100] (* _Jean-François Alcover_, Nov 12 2020, after _Alois P. Heinz_ *)
%Y Cf. A140358.
%K nonn
%O 1,2
%A _John W. Layman_, Oct 18 2011
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