A197702 Smallest positive integer k such that n = +-1 +-3 +-... +-(2k-1) for some choice of +'s and -'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, 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, 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

Offset: 1

Author

John W. Layman, Oct 18 2011

Keywords

• nonn

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)

Examples

			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.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A140358.

Programs

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 *)

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