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I am afraid to continue to ask trivial things but really I do not know how to proceed, so I ask the experts:

A specially multiplicative function, is a function $f$ from positive integers to complex numbers, that satisfies: $$ f(n)f(m) = \sum_{d \mid \gcd(m,n)} f(\frac{mn}{d^2})g(d) $$ for all $m,n$, where $g$ is a completely multiplicative function, i;e. $$ g(rs) = g(r)g(s) $$ for all positive integers $r,s.$

Seems well known that any specially multiplicative function $F$ is the Dirichlet convolution of $2$ completely multiplicative functions. (We write, say, $F = r * s$). That is:

$$ F(n) = r * s (n) = \sum_{d \mid n} r(d) s(n/d) $$ with $r,s$ completely multiplicative functions.

I checked that

$$ \sigma = id * z $$

where $\sigma$ is the sum of divisors's function, (that is specially multiplicative with a explicit $g$), $$ id(n) = n $$ for all $n$ and $$ z(n) = 1 $$ for all $n.$

But I am (unfortunately) unable to find completely multiplicative functions $a,b$ such that

$$ \tau = a * b $$

where $\tau$ is Ramanujan's tau function.

Question 1 : Somebody can display such $a,b$ ???

Probably, there is a standard construction of such $a,b$ for any specially multiplicative function.

Question 2 : What is the recipe to get such $a,b$ for a given specially multiplicative function ???

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I don't understand your definition of "specially multiplicative"; if $m$ and $n$ are coprime, it gives $f(m)f(n)=mn$, which is not true of the sum of divisor function...

However, the Ramanujan $\tau$-function satisfies $\tau(m)\tau(n)=\sum_{d\mid (m,n)}{\mu(d)\tau(mn/d^2)}$, which might be what you mean?

In any case, $\tau$ is indeed the Dirichlet convolution of two totally multiplicative functions: factor the local $p$-factor of the Dirichlet generating series, which is $1-\tau(p)p^{-s}+p^{11-2s}$, as $(1-a(p)p^{11/2-s})(1-b(p)p^{11/2-s})$, then the totally multiplicative functions mapping $p$ to $a(p)$ and $b(p)$ satisfy $a* b=\tau$.

There is therefore (uncountably) many possibilities for these functions (with $a(p)$ being chosen out of two possibilities for every $p$), so it is hard to get a simple formula (especially a formula that does not involve factoring $n$ into primes).

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  • $\begingroup$ Apologies! : I forgetted the "f" in the right hand side. Would edit and correct this. Thanks a lot for the answer. For question 2, is the same thing ? $\endgroup$ – Luis H Gallardo Jun 13 '11 at 20:24
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Paul McCarthy states in the notes of chapter 1 of his book "Introduction to Arithmetical Functions" that the completely multiplicative functions $g_1,g_2$ such that $\tau=g_1*g_2$ are defined on primes in the following way: $$g_1(p)=\frac{1}{2}\left(\tau(p)+\sqrt{\tau(p)^2-4p^{11}}\right)$$ $$g_2(p)=\frac{1}{2}\left(\tau(p)-\sqrt{\tau(p)^2-4p^{11}}\right)$$

(You can also find the general recipe in chapter 1 of the book)

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