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Conventions

We will call the following product the canonical expansion of $a$: $$a = \prod\limits_{p_i\in\mathbb{P}}p_i^{\alpha_i}.$$

If $a$ and $b$ are co-prime we write $$a \perp b.$$

$\beth$-function

Suppose $a\in \mathbb{N}_+$ and $M_k(a)$ is the following set: $$M_k(a) = \Big\{ (x_1,\ldots,x_k) \Big| \sum x_i = a\quad \mathrm{and}\quad x_i\perp x_j\quad \mathrm{for}\quad i \neq j\Big\}.$$ Then let us construct function (\beth): $$\beth_k(a) = |M_k(a)|.$$ We choose by definition $$\beth_k(1) = 1.$$ In other words, $\beth_k(a)$ counts co-prime $k$-partitions of $a$.

So my question is to investigate $\beth$-function's properties. The main question, no doubt, is

Write the explicit (closed form) expression for $\beth_k(a)$ using the canonical expansion of $a$.

Now I am trying to prove that $\beth_k$ is a multiplicative function. More specific, I want to prove that $$\beth_k(p_1p_2) = \beth_k(p_1)\beth_k(p_2),\quad\forall p_1, p_2\geq k,\, p_1 \perp p_2.$$

Improve this question
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  • $\begingroup$ Some empirical observations which might or might not be helpful or deceptive$$\beth_k(k)=1$$ $$\beth_k(k+1)=\beth_k(k+2)=\beth_k(k+4)=k$$ $$\beth_k(k+3)=\beth_k(k+6)=\beth_k(k+8)=k^2$$ $$\beth_k(k+5)=2k^2-k$$ $$\beth_k(k+7)=k^3-k^2+k$$ $$\beth_k(k+9)=2k^3-2k^2+k$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 10, 2017 at 8:33
  • $\begingroup$ (contd.)$$\beth_k(k+10)=k^2$$$$\beth_k(k+11)=2k^3-k^2$$$$\beth_k(k+12)=k^3-k^2+k$$$$\beth_k(k+13)=k^4-2k^3+2k^2$$$$\beth_k(k+14)=3k^2-2k$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Feb 10, 2017 at 8:51
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    $\begingroup$ One has $\beth_{pq}(pq) = 1 \neq 0 = \beth_{pq}(p) \beth_{pq}(q)$ for distinct primes $p,q$. Thus $\beth_{k}$ is not multiplicative in general. $\endgroup$
    – js21
    Commented Feb 10, 2017 at 9:38
  • $\begingroup$ @js21, Of course $\beth_k(a)$ is define for only $a\geq k$. I will add your remark to the description. P.S, I have tested $\beth_2(n)$ for multiplicity for $n$ up to 100. $\endgroup$
    – LRDPRDX
    Commented Feb 10, 2017 at 9:56
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    $\begingroup$ $\beth_2$ is just Euler's totient function. The case $k=2$ is a very special case. $\endgroup$
    – js21
    Commented Feb 10, 2017 at 10:01

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