Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that for two integral-valued arithmetic functions $f_i$ ($i=1,2$), the following values are known: $$ \lim_{n\to \infty} \frac{ \{n:f_i(n) \text{ is odd}\}\cap \{0,1,\ldots, n-1\} }{n}.$$ (In other words the density of odd $f_i(n)$'s are known.)

Is there any result which relates these densities to the following density: $$ \lim_{n\to \infty} \frac{ \{n: \sum_{i=0}^{n} f_1(i)f_2(n-i) \text{ is odd}\}\cap \{0,1,\ldots, n-1\} }{n}?$$

Note that the sum inside occurs when multiplying the generating functions for $f_1$ and $f_2$. Indeed, $$ (\sum_{n=0}^{\infty} f_1(n)q^n)(\sum_{n=0}^{\infty} f_2(n)q^n) = (\sum_{n=0}^{\infty} (\sum_{i=0}^{n} f_1(i)f_2(n-i))q^n).$$

This occurred to me while studying the parity of certain functions.

share|improve this question
    
Let's denote $D(f)$ ( or more generally, $D_*(f)$ and $D^*(f)$) the density of the support of the sequence $f$ $\operatorname{mod} 2$, that you are considering (resp. the lower and upper densities). I may be wrong, but my feeling is that the result is, that there is no result, meaning that, given $f_1$ and $f_2$, you can perturb them with sequences $h_1, h_2$ with zero-densities (I mean $D(h_1)=D(h_2)=0$), and obtain any prescribed $D_*( (f_1+h_1)*(f_2 + h_2))$, and $D^*( (f_1 + h_1) * (f_2 + h_2))$. Of course, $D_*(f_i+h_i)=D_*(f_i)$ and $D^*(f_i+h_i)=D^*(f_i)$. –  Pietro Majer Feb 21 '12 at 15:30
    
In other words, I suspect that if one knows nothing but the first two densities, nothing can be said on the third. –  Pietro Majer Feb 21 '12 at 18:05
    
You are right, Majer. The two densities are not enough to conclude anything about the third one. I tried the problem for the sequence $f=(0,0,0,1,0,0,0,1,\ldots)$ with density 1/4, together with each of the sequences $g_1=(0,1,0,1,0,1,0,1,\ldots)$ and $g_2=(0,0,1,1,0,1,1,0,\ldots)$ with densities 1/2, and obtained two different densities for their convolution. Thank you for the suggestion –  Victor Aricheta Feb 26 '12 at 7:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.