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{\Large Rational Function Multiplicative Coefficients} \\
Michael Somos 11 Dec 2014 \\
ms639@georgeown.edu \\
(draft version 11) \\
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\section{Rational generating functions of multiplicative sequences}
Any numerical sequence has an associated generating function (GF). For example,
the Fibonacci sequence is associated with GF $x / (1 - x - x^2)$, a rational
function of $x$. Consider a multiplicative sequence. That is, $a(1) = 1$ and
$a(n m) = a(n) a(m)$ for all positive integers $n$ and $m$ relatively prime
to each other. Can its GF $f(x) = a(1) x + a(2) x^2 + a(3) x^3 + ...$ ever be
rational? The answer is yes if $f(x) = x / (1 - x)$ and $a(n) = 1$ if $n > 0$.
This is the simplest example where $a(n)$ is non-zero for all $n > 0$. Another
is $f(x) = x / (1 - x^2)$ and $a(n) = 1$ if $n > 0$ is odd and $a(n) = 0$
otherwise. Now consider the rational function and its power series expansion
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$$f(x) = x (1 - x)^{e_1} (1 - x^2)^{e_2} = x - e_1 x^2 + ((e_1^2 - e_1)/2 - e_2) x^3 + ...$$
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where $e_1$ and $e_2$ are integers. A search finds that $f(x)$ is the GF of a
multiplicative sequence for 11 pairs of integers $[e_1,e_2]$ as follows:
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$$[-4,1],[-2,0],[-1,0],[-1,1],[0,-1],[0,0],[0,1],[1,-1],[1,0],[2,-2],[4,-3].$$
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The multiplicative integer sequences for these pairs are of a simple form. Some
algebra is enough to prove that this list is complete. Allowing more factors in
$f(x)$ increases the difficulty of search and algebraic proof.
\section{Conjecture 1}
Conjecture 1: there is a finite set of rational functions of the form
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$$ f(x) = x (1 - x)^{e_1} (1 - x^2)^{e_2} ... (1 - x^n)^{e_n} $$
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for some integers $e_1, ..., e_n$ which are the GF for multiplicative integer
sequences provided we exclude some infinite families which are predictable.
One example infinite family is
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$$ f(x) = x (1 - x^{n-1}) = x - x^n $$
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where $n > 1$. Also, if and only if $n = p^k, n > 1$ and $p$ is prime then
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$$ f(x) = x (1 - x)^{-1} (1 - x^{n-1}) (1 - x^n)^{-1} $$
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is multiplicative. Note that $f(x)$ is in the set when $-f(-x)$ is since
$(1 + x) = (1 - x^2) / (1 - x)$ and so on.
\section{Homogeneous generalization of multiplicative sequences}
Now assume $a(0)$ and $a(1)$ are nonzero. For example, consider the sequence
$a(n)$ with its GF
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$$ g(x) = (1 - x)^2 / (1 - x^2) = 1 - 2 x + 2 x^2 - 2 x^3 + ... . $$
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Then $a(1) a(n m) = a(n) a(m)$ for all positive integers $n$ and $m$ relatively
prime to each other. This is a homogeneous generalization of multiplicative
sequences. As in the first section, but without a factor of $x$, consider
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$$g(x) = (1 - x)^{e_1} (1 - x^2)^{e_2} = 1 - e_1 x + ((e_1^2 - e_1)/2 - e_2) x^2 + ...$$
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where $e_1$ and $e_2$ are integers. A search finds that $g(x)$ is the GF of a
homogeneous multiplicative sequence for 10 pairs of integers $[e_1,e_2]$
as follows:
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$$[-4,2],[-2,1],[-2,2],[-1,0],[-1,1],[1,-1],[1,0],[2,-1],[2,0],[4,-2].$$
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Again, algebra is enough to prove the list is complete.
\section{Conjecture 2}
Conjecture 2: there is a finite set of rational functions of the form
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$$ g(x) = (1 - x)^{e_1} (1 - x^2)^{e_2} ... (1 - x^n)^{e_n}$$
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for some integers $e_1, ..., e_n$ which are the GF for homogeneous
multiplicative integer sequences provided we exclude some infinite families
which are predictable. For example,
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$$ g(x) = (1 - x)^{-1} (1 - x^n)^{-1} (1 - x^{n+1})$$
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is homogeneous multiplicative if and only if $n = p^k, n > 1$ and $p$ is prime.
Note that $g(x)$ is in the set when $g(-x)$ is.
\section{Further Work}
The rational functions in the two conjectures have applications related to
Ramanujan's Lambert series. A study of rational functions with poles only at
roots of unity appeared in 2003 by Juan B. Gil and Sinai Robins who defined a
Hecke operator on power series. Kyoji Saito studied cyclotomic functions
related to eta-products in 2001. Rational functions of a simple form having
multiplicative coefficients is related to a paper on Multiplicative $\eta$-
Quotients by Yves Martin in 1996.
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