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.

Another Result

\[\lim_{N\to\infty}\left(\sum_{k=1}^N x^{x^{p_k}}\right)-N=\sum_{q=2}^\infty \left(\sum_{p|q} \frac{\log(x)^{q/p}}{\left(\frac{q}{p}\right)!}\right)x^q\\ \lim_{N\to\infty}\left(\sum_{k=1}^N x^{x^{p_k}}\right)-N= \sum_{p\in\mathbb{P}} \sum_{q=1}^\infty \frac{\log(x)^qx^{qp}}{q!}\]

where \(p\) is a prime, and \(p_k\) is the \(k^{th}\) prime. This implies \[\sum_{q=2}^\infty \left(\sum_{p|q} \frac{\log(x)^{q/p}}{\left(\frac{q}{p}\right)!}\right)x^q=\sum_{p\in\mathbb{P}} \sum_{q=1}^\infty \frac{\log(x)^qx^{qp}}{q!}\] which appears to revolve around a substitution of \(q\to q/p\)

we can then suggest that if \[\sum_{q=1}^\infty \sum_{p|q} \frac{p}{q}x^q(1-x^q) = \log\left(\prod_{k=1}^\infty 1+x^{p_k} \right)\] then \[\sum_{q=1}^\infty \sum_{p\in\mathbb{P}} \frac{1}{q}x^{qp}(1-x^{qp}) = \log\left(\prod_{k=1}^\infty 1+x^{p_k} \right)\] which appears to be true by looking at the expansions. If we do something of the opposite to the transform we get \[\sum_{p\in\mathbb{P}} x^p - x^{2p}\]

But we have generated a transform for \[T\sum_{p}f(p) \to \sum_{q=1}^\infty \sum_{p} \frac{f(qp)}{q}\] such that \[T\sum_{p\in\mathbb{P}} x^p \to \sum_{q=1}^\infty \sum_{p\in\mathbb{P}} \frac{x^{p q}}{q} = \sum_{k=2}^\infty \frac{\mathrm{sopf}(k)x^k}{k}\] where \(\mathrm{sopf}(k)\) is the sum of the prime factors of \(k\). \[\sum_{q=1}^\infty \sum_{k=1}^\infty \frac{x^{p_{q k}}}{q} =\sum_{k=1}^\infty \frac{\sigma_1(k)x^{p_k}}{k}\] It would appear that \[T\sum_{p\in\mathbb{P}} x^{p^2} \to \sum_{q=1}^\infty \sum_{p\in\mathbb{P}} \frac{x^{p^2 q}}{q}\]