

A197702


Smallest positive integer k such that n = +1 +3 +... +(2k1) for some choice of +'s and 's


2



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
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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(k1)<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



EXAMPLE

The sum of 3 terms 1  3 + 5 gives 3, but none of the 2term sums 1+3, 13, 1+3, 13 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(n2*i+1, i1) or b(n+2*i1, i1))
end:
a:= proc(n) local k;
for k from floor(sqrt(n)) while not b(n, k) do od; k
end:


MATHEMATICA

b[n_, i_] := b[n, i] = (n==0 && i==0)  Abs[n] <= i^2 && (b[n2i+1, i1]  b[n+2i1, i1]);
a[n_] := Module[{k}, For[k = Floor[Sqrt[n]], !b[n, k], k++]; k];


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



