# Integrable continuations of an arithmetic function.

Suppose we have the arithmetic functions $a_i : \mathbb{N} \rightarrow \mathbb{R}$ and $b_j : \mathbb{N} \rightarrow \mathbb{R}$ such that $a_n = b_n = 1$ if and only if the index $n$ is a square, and $a_n = b_n = 0$ otherwise. Suppose now that we would like to analyze $\sum_{i \ge 1} {a_i x^i} \sum_{j \ge 1} {b_j x^i}$ without convolutions to begin with. The approach is to extend $a_i$ and $b_j$ with continuous interpolations such that an integral rather than a sum might be evaluated.

That is, we want to create continuous auxiliary functions $a^*_i : \mathbb{R} \rightarrow \mathbb{R}$ and $b^*_j : \mathbb{R} \rightarrow \mathbb{R}$ so that $\forall m \in \mathbb{N}, a^*_m = a_m$ and $b^*_m = b_m$. The hope is that we can now enforce the condition on our $a^*_n, b^*_n$ so that $a^*_i b^*_j + a^*_j b^*_i = 0$ when $i, j \not \in \mathbb{N}$ and $a^*_i b^*_j + a^*_j b^*_i = a_i b_j + a_j b_i$ when $i, j \in \mathbb{N}$ and evaluate the double integral $\int_{i \gt 0} {a^*_i x^i} \int_{j \gt 0} {b^*_j x^j} = \int_{i \gt 0} {\int_{j \gt 0} a^*_i b^*_j x^{i+j}}$ into a target function with support $\mathbb{R} \rightarrow \mathbb{R}$. The goal then, would be to use fourier analysis on this integral to collect coefficients without the usual arithmetic convolution.

Can one prove from the conditions states that no such auxiliary functions exist?

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Hopefully my description is clear enough. It may be that more details are needed. – Ross Snider Feb 20 '13 at 23:48
Sometimes, it seems you have a sequence $a_1(n),a_2(n),\dots$ of arithmetic functions; sometimes, it seems you have a single arithmetic function $a$ whose values are the numbers $a_1,a_2,\dots$, that is, $a(i)=a_i$. If it's the latter, then you want $a_4^*b_9^*+a_9^*b_4^*=2$ but $a_i^*b_j^*+a_j^*b_i^*=0$ for $i,j$ arbitrarily close to $4,9$, and there goes continuity. – Gerry Myerson Feb 21 '13 at 0:32
It's the latter. That easy, huh? Thanks Gerry! :-D – Ross Snider Feb 21 '13 at 1:22