OFFSET
1,2
COMMENTS
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.)
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..10000
EXAMPLE
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.
MAPLE
b:= proc(n, i) option remember; (n=0 and i=0) or
abs(n)<=i^2 and (b(n-2*i+1, i-1) or b(n+2*i-1, i-1))
end:
a:= proc(n) local k;
for k from floor(sqrt(n)) while not b(n, k) do od; k
end:
seq(a(n), n=1..100); # Alois P. Heinz, Oct 19 2011
MATHEMATICA
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]);
a[n_] := Module[{k}, For[k = Floor[Sqrt[n]], !b[n, k], k++]; k];
Array[a, 100] (* Jean-François Alcover, Nov 12 2020, after Alois P. Heinz *)
CROSSREFS
KEYWORD
nonn
AUTHOR
John W. Layman, Oct 18 2011
STATUS
approved