When is A111075(n) Odd?

[Max]

Paul Barry wrote:
> Seqfans,
> 
> With regard to this discussion, note that the Jacobsthal case :
> 
> Trading 
> A111075 : F(n) * sum{k|n} 1/F(k), where F(k) is the kth Fibonacci number
> 
> for 
> 
> J(n) * sum{k|n} 1/J(k), where J(k) is the kth Jacobsthal number (A001045)
> 
> leads to a set of numbers a(n) that are odd precisely for n=m^2?

That's trivial.
Since J(m) is odd for m>0 and J(k) divides J(n) as soon as k divides n, we have

sum{k|n} J(n)/J(k) mod 2 = sum{k|n} 1 mod 2 = numdiv(n) mod 2

which odd iff n is a square.

Max