# An Infinite Product Expansion

Let $\Pi_N(a,x)= \prod_{k=1}^N a+x^{p_k}$ Where $$p_k$$ is the $$k^{th}$$ prime. Then $\lim_{N\to\infty}(\log\Pi_N(a,x) - \log a^N) = \frac{2x^2}{2a}+\frac{3x^3}{3a} -\frac{2x^4}{4a^2} + \frac{5x^5}{5a} + \frac{(2-3a)x^6}{6a^3} +\frac{7x^7}{7a} -\frac{2x^8}{8a^4}+\frac{3x^9}{9a^3}+\frac{(2-5a^3)x^{10}}{10a^5}+\cdots$ It would appear there are a few patterns here.

• The power of $$a$$ in the denominator is the largest proper divisor of the power of $$x$$.

• The denominator is the power of $$x$$.

• For prime powers of $$x$$ the numerator is also the power of $$x$$.

• For a perfect square or cube power of $$x$$, the prime divisor is the numerator.

• For a composite power of $$x$$ the numerator is a polynomial in $$a$$, with coefficients of the divisors of $$x$$.

• After removing the denominator coefficient of the power of $$x$$, all raw coefficients present in the above expansion are prime.

Focus on the composites $x^4 \to -\frac{2}{a^2}\\ x^6 \to \frac{2-3a}{a^3}\\ x^9 \to \frac{3}{a^3}\\ x^{10} \to \frac{2-5a^3}{a^5}\\ x^{12} \to -\frac{2+3a^2}{a^6}\\ x^{42} \to \frac{2-3a^7-7a^{15}}{a^{21}}\\ x^{44} \to -\frac{2+11a^{18}}{a^{22}}\\ x^{30} \to -\frac{-2+3a^5+5a^9}{a^{15}}\\ x^{66} \to -\frac{-2+3a^{11}+11a^{27}}{a^{33}}\\ x^{69} \to \frac{3+23a^{20}}{a^{23}}$ It seems that for numbers with $$3$$ prime factors, the coefficient of the largest power of $$a$$ in the numerator is the power of the second largest power of $$a$$. Although only a few numbers have been checked so far.

It would appear that for a polynomial of the form $x^q \to \frac{x+ya^m+za^n}{a^o}\\ xo=q\\ y(o-m)=q\\ z(o-n)=q$ giving $x^q \to \frac{1}{a^{D_m(q)}}\sum_{p|q} (-1)^?pa^{D_m(q)-\frac{q}{p}}$ where $$D_m(q)$$ is the largest proper divisor of $$q$$. This gives $\lim_{N\to\infty}(\log\Pi_N(a,x) - \log a^N) = \sum_{q=2}^\infty \frac{1}{qa^{D_m(q)}}\sum_{p|q} (-1)^?pa^{D_m(q)-\frac{q}{p}}x^q \\ \prod_{k=1}^\infty 1+x^{p_k}=\exp\left(\sum_{q=2}^\infty \frac{1}{q}\sum_{p|q} (-1)^?px^q\right)$

where the pattern for the negative signs is yet to be found. It would appear that if the power of $$x$$ is divisible by $$2$$ and the power of $$x$$ divided by $$2$$ is not equal to $$1$$, then the overall sign of the temr in the expansion is negative. Then the polynomials in the numerator must be investigated. It would appear that the polynomial coefficient of $$2$$ is negative if the power of $$x$$ is of the form $$4n+2$$, $$n=0,1,2,3,...$$. All other prime divisor polynomial coefficients appear to be positive. This then explains all terms (as far as we know). This gives $\lim_{N\to\infty}(\log\Pi_N(a,x) - \log a^N) = \sum_{q=2}^\infty \frac{1}{qa^{D_m(q)}}\sum_{p|q} (-1)^{q+1}p^*(q)a^{D_m(q)-\frac{q}{p}}x^q \\ \prod_{k=1}^\infty 1+x^{p_k}=\exp\left(\sum_{q=2}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)$ where $$p^*(q)$$ means $p^*(q)=(-1)^{q/2}\cdot 2,\; p=2\\ p^*(q)=p,\; \mathrm{otherwise}$ We may now manipulate this expression $\prod_{k=1}^\infty 1+x^{p_k}=\exp\left(\left[\sum_{q\in\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right] + \left[\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right]\right) \\ \prod_{k=1}^\infty 1+x^{p_k}=\exp\left(\sum_{q\in\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)\exp\left(\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)\\ \frac{\prod_{k=1}^\infty 1+x^{p_k}}{\exp\left(\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)}=\exp\left(\sum_{q\in\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)\\ \log\left(\frac{\prod_{k=1}^\infty 1+x^{p_k}}{\exp\left(\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)}\right)=\sum_{q\in\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\\ \log\left(\frac{\prod_{k=1}^\infty 1+x^{p_k}}{\exp\left(\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)}\right)=\sum_{q\in\mathbb{P}}^\infty x^q$ Where the right hand side is now the ordinary generating function for primes. Then of course we also have $\log\left(\frac{\prod_{k=1}^\infty 1+x^{p_k}}{\exp\left(\sum_{q\in\mathbb{P}}^\infty x^q\right)}\right)=\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q$ Then if we set $$x=1$$, and change the infinite sums to $$n$$ (checked that the series are generated at the same rate) we should have have the prime counting function $\pi(n) = \left.\log\left(\frac{\prod_{k=1}^\infty 1+x^{p_k}}{\exp\left(\sum_{q\notin\mathbb{P}}^\infty \frac{1}{q}\sum_{p|q} (-1)^{q+1}p^*(q)x^q\right)}\right)\right|_{x=1}$

# Another Form

$\log\left(\prod_{k=1}^\infty q+x^k\right) = \frac{(-1+2a)x^2}{2a^2} + \frac{(1+3a^2)x^3}{3a^3}+\frac{(-1-2a^2+4a^3)x^4}{4a^4}+\frac{(1+5a^4)x^5}{5a^5}+\frac{(-1+2a^3-3a^4+6a^5)x^6}{6a^6}+\cdots$

So we clearly have $\log\left(\prod_{k=1}^\infty q+x^k\right) = \sum_{q=2}^\infty \frac{x^q\sum_{d|q}(-1)^?da^{q(1-1/d)}}{qa^q}$

We can express a general form. Define $$A$$ to be a token $A=\begin{bmatrix}0& 1\\0 &0 \end{bmatrix}$ such that $$A^k=\emptyset$$, for $$k>1$$. Then $\log\left(\prod_{k=1}^\infty Ax^k+f(k)\right) = A\sum_{k=1}^\infty \frac{x^k}{f(k)}$ For example Then $\log\left(\prod_{k=1}^\infty Ax^k+k!\right) = A(\exp(x)-1)$ Then define a token $$B$$ such that $$B^k=0$$, $$k>2$$. $\log\left(\prod_{k=1}^\infty Bx^k+f(k)\right) = B\sum_{k=1}^\infty \frac{x^k}{f(k)} -\frac{B^2}{2}\sum_{k=1}^\infty \frac{x^{2k}}{f(k)^2}$ and so on till we have an infinite token $$Z$$, or just a constant, and $\log\left(\prod_{k=1}^\infty Zx^k+f(k)\right) = Z\sum_{k=1}^\infty \frac{x^k}{f(k)} -\frac{Z^2}{2}\sum_{k=1}^\infty \frac{x^{2k}}{f(k)^2} +\frac{Z^3}{3}\sum_{k=1}^\infty \frac{x^{3k}}{f(k)^3}-\frac{Z^4}{4}\sum_{k=1}^\infty \frac{x^{4k}}{f(k)^4} +\cdots\\ \log\left(\prod_{k=1}^\infty Zx^k+f(k)\right) = \sum_{j=1}^\infty \frac{(-1)^{j+1}Z^j}{j}\sum_{k=1}^\infty \frac{x^{jk}}{f(k)^j}\\ \log\left(\prod_{k=1}^\infty Zx^k+f(k)\right) = \log\left(1+Z\tau\right)\\$ The last equation being on the condition that $\sum_{k=1}^\infty \frac{x^{jk}}{f(k)^j} = \tau^j$ which is probably impossible... However if we set $f(k)\to f(k,s)=k^sx^k$ we get $\log\left(\prod_{k=1}^\infty Zx^k+k^sx^k\right) = \sum_{j=1}^\infty \frac{(-1)^{j+1}Z^j}{j}\sum_{k=1}^\infty \frac{1}{k^{js}}\\ \log\left(\prod_{k=1}^\infty Zx^k+k^sx^k\right) = \sum_{j=1}^\infty \frac{(-1)^{j+1}Z^j}{j} \zeta(js)\\ \log\left(\prod_{k=1}^\infty Zx^k+kx^k\right) = Z\zeta(1) -Z\gamma - \log(\Gamma(1+Z))\\$

# One Strange Result

$\sum_{q=1}^\infty \frac{1}{q}\sum_{p|q} px^q - \log\left(\prod_{k=1}^\infty 1+x^{p_k} \right) = \sum_{q=1}^\infty \frac{1}{q}\sum_{p|q} px^{2q}$

then $\sum_{q=1}^\infty \frac{1}{q}\sum_{p|q} px^q(1-x^q) = \log\left(\prod_{k=1}^\infty 1+x^{p_k} \right)$ we also have $\sum_{q=1}^\infty \frac{1}{q}\sum_{d|q} dx^q(1-x^q) = \log\left(\prod_{k=1}^\infty 1+x^{k} \right)$ these are then statements that $(1+x^2)(1+x^3)(1+x^5)(1+x^7)(1+x^{11})(1+x^{13})\cdots=\exp\left(\frac{1}{2}(2x^2-2x^4)+\frac{1}{3}(3x^3-3x^6)+\frac{1}{4}(2x^4-2x^8)+\frac{1}{5}(5x^5-5x^{10})+\cdots \right)$ we also seem to have $\sum_{q=1}^\infty \frac{1}{q}\sum_{s|q} sx^q(1-x^q) = \log\left(\prod_{k=1}^\infty 1+x^{s_k} \right)$ where $$s$$ is a square free number, and $$s_k$$ is the $$k^{th}$$ square free number.