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\begin{document}

\title{Further Sequence Transforms, Sum over infinite set of indices conforming to a rule}



\author[1]{Benedict Irwin}%
\affil[1]{University of Cambridge}%


\vspace{-1em}



  
  \date{May 22, 2024}


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\sloppy


\section{Abstract}
I consider defining transforms to mimic the inverse series transform in structure
\section{Main}
If we have an arbitrary series expansion,
\begin{equation}
P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = \sum_{k=0}^\infty a_kx^k
\end{equation}
then we can describe the inverse term by \begin{equation}
P^{-1}(x) = \sum_{n=1}^\infty \frac{1}{na_1^n}\left(\sum_{s,t,u,\cdots}(-1)^{s+t+u+\cdots}\frac{n(n+1)\cdots(n-1+s+t+u\cdots)}{s!t!u!\cdots}\left(\frac{a_2}{a_1}\right)^s\left(\frac{a_3}{a_1}\right)^t\cdots\right)(x-a_0)^n
\end{equation}
[Morse and Feshbach (1953)], where, critically $s+2t+3u+\cdots=n-1$, all of $s,t,u,\cdots$ are positive integers and the sum is then over the combinations to do this.
This transform is clearly very important, can we create a similar looking transform, and see if any integer sequences are related by the two?
\section{Example 1}
Again let
\begin{equation}
P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = \sum_{k=0}^\infty a_kx^k
\end{equation}
this time define the transformed polynomial as \begin{equation}
\mathcal{T_1}[P](x) = \sum_{n=1}^\infty \left(\sum_{s,t,u,\cdots}(sa_1+ta_2+ua_3+\cdots)\right)(x-a_0)^n\\
s+t+u+\cdots=n \\
s\ge t \ge u \ge \cdots
\end{equation}
we may need to enforce that the higher terms can be $0$, the definition can be reworked if this is the case. Then we end up with \begin{equation}
\mathcal{T_1}[P](x) = (a_1)(x-a_0)+(3a_1+a_2)(x-a_0)^2+(6a_1+2a_2+a_3)(x-a_0)^3+(12a_1+5a_2+2a_3+a_4)(x-a_0)^4 + \cdots
\end{equation}
which looks like it may well have a nice form for certain input $a_k$ terms. If we use $P(x)=x$, then we find that the transform is the generating function for A006128. \begin{equation}
\mathcal{T_1}[x] = \sum_{n>0} \frac{nx^n}{\prod_{k=1}^n(1-x^k)}
\end{equation}
likewise for $P(x)=x^2$, we have the generating function for A096541.\\
For $P(x)=x+x^2+x^3+x^4+\cdots$, we get the generating function for A066186.\\
For $P(x)=x-x^2+x^3-x^4+\cdots$, we get the generating function for A066897.\\
For $P(x)=x+2x^2+3x^3+4x^4+\cdots$, we get the generating function for A066184.\\
For $P(x)=x+x^3+x^5+\cdots$, we get A207381.\\
To generate the generating function for the natural numbers we need to put in the function $F_N(x)=x -x^2 -x^3 -x^4 -x^5 +x^6 -x^7 +3x^8 + 0x^9 + \cdots$, the rule is not clear.\\
Interestingly if we input $P(x)=F_N(x)+x$, we get the generating function for A225596.\\
To generate the generating function for the prime numbers we need to put in the function $F_P(x)=2x -3x^2 -x^3 -x^5 +x^6 -4x^7 +3x^8 +0x^9 + \cdots$, the rule is not clear.
If we input $P(x)=F_P(x)-F_N(x)$, we do seem to get the generating function for A014689, which makes sense.\\ 
To get $\pi(n)$, we can use the coefficients $1,-1,-2,0,-1,3,-2,2,2,\cdots$

\section{Change Inequalities}
If we force the inequalities to be strict, that is $s> t > u > \cdots$,\\
then with $P(x)=x$ we have the g.f. of A005895.
with $P(x)=x+x^2+x^3+x^4+\cdots$, we have the g.f. of A066189.

\section{Type 2}
What about the following transform \begin{equation}
P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = \sum_{k=0}^\infty a_kx^k
\end{equation}
then we can describe the inverse term by \begin{equation}
T_2[P(x)] = \sum_{n=1}^\infty \left(\sum_{s,t,u,\cdots}(-1)^{s+t+u+\cdots}a_0^sa_1^ta_2^ua_3^w\cdots\right)x^n\\
s+2t+3u+\cdots=n-1
\end{equation}
Does this generate anything coherent? We get \begin{equation}
T_2[P(x)] = x - a_0x^2 + (a_0^2 - a_1)x^3 + (a_0a_1 - a_0^3 - a_2)x^4 + (a_0a_2 + a_1^2 - a_3 - a_0^2a_1 + a_0^4)x^5+\cdots
\end{equation}
with $P(x)=1$, we will then get \begin{equation}
T_2[1] = x - x^2 + x^3 - x^4 + x^5 + \cdots
\end{equation}
with $P(x)=1+x$, we will then get \begin{equation}
T_2[1+x]= x - x^2 + x^5 -x^6 + \cdots = \sum_{k=0}^\infty x^{4k+1}-x^{4k+2}
\end{equation}
with $P(x)=1+x^2$, we will then get \begin{equation}
T_2[1+x]= x-x^2+x^3-2x^4+2x^5-2x^6+3x^7-3x^8+3x^9-4x^{10}+\cdots
\end{equation}
nicely if we input $P(x)=1+2x+3x^2+4x^3+\cdots$, then we have \begin{equation}
T_2\left[\frac{1}{(x-1)^2}\right]= x -x^2 -x^3 -2x^4 +2x^5 -x^6 + 4x^7 -x^8 + 18x^9 -22x^{10} + \cdots\\
T_2\left[\frac{1}{(x-1)^2}\right]= x \prod_{m>0} \frac{1}{1+mx^m}
\end{equation}
where the bottom equaility is due to A022693.
It would also seem that \begin{equation}
T_2[1+2x]= \frac{x}{1+x+2x^2+2x^3}
\end{equation}
for \begin{equation}
T_2[1-x+x^2-x^3+x^4-\cdots]
\end{equation}
we have the alternating signs of the partition numbers, A000041.
\section{Possible Results}
Below I list a few significant transformations of simple input generating functions. There appears to be a strong connection to modular forms.
\begin{equation}
T_2\left[\frac{1}{(x-1)^2}\right]= x \prod_{m>0} \frac{1}{1+mx^m}\\
T_2\left[\frac{3}{x-1}\right]=\frac{x}{\prod_{m>0} (1-3x^m)}\\
T_2\left[\frac{1}{1-x}\right]= \frac{x^{25/24}\eta(t)}{\eta(t^2)}\\
T_2\left[\frac{1}{1+x}\right]= \frac{(-1)^{1/24}x^{25/24}}{\eta(-\tau)}\\
T_2\left[\frac{1}{x^2-1}\right]=\frac{x^{23/24}\eta(t^2)}{\eta(t)}\\
T_2\left[\frac{1}{1-x^2}\right]= x^{23/24}\left(\frac{m(1-m)}{16}\right)^{1/24}\\
T_2\left[\frac{1+x}{1-x^3}\right] = \frac{x^{11/12}\eta(t)\eta(t^6)}{\eta(t^2)\eta(t^3)}\\
T_2\left[\frac{1+x}{x^3-1}\right] = \frac{x^{11/12}\eta(t^3)}{\eta(t)}\\
T_2\left[\frac{x}{x^4-1}\right] = \frac{x^{11/12}\eta(t^4)}{\eta(t^2)}\\
T_2\left[1+x+x^4-x^5+x^6-x^7+x^9+x^{10}-x^{11}+x^{12}+x^{13}-x^{15}+x^{16}-x^{17}+\cdots\right] =  \frac{x^{5/6}\eta(t)\eta(t^{12})}{\eta(t^2)\eta(t^3)\eta(t^4)}\\
T_2[0 0 -1 1 0 -1 0 2 -1 0 0 0 0 0 -1 4 0 -1] = \frac{x^{23/24}\eta(t^4)}{\eta(t^3)}\\
T_2\left[ \frac{x+x^3}{x^6-1} \right] = \frac{x^{5/6}\eta(t^6)}{\eta(t^2)}\\
T_2\left[ \frac{x^2}{x^6-1} \right] = \frac{x^{7/8}\eta(t^6)}{\eta(t^3)}\\
t^n=\frac{n\log(x)}{2\pi i}\\
\tau^n=\frac{n\log(-x)}{2\pi i}\\
m=\lambda\left(t^2\right)
\end{equation}
with $\eta(q)$, Dedekind Eta function, $\lambda$, is the modular lambda function.

\section{How Does This Come About}
We can observe a similar set of sums, with restricted tuples of indices. \begin{equation}
f_1(k)=\sum_{i_1:i_1=k}i_1 \to 1,2,3,4,5,6,7,\cdots\\
f_2(k)=\sum_{i_1,i_2:i_1+2i_2=k} i_1i_2 \to 1,2,5,8,14,20,30,\cdots\\
f_3(k)=\sum_{i_1,i_2,i_3:i_1+2i_2+3i_3=k} i_1i_2i_3 \to 1,2,5,10,18,30,49,\cdots
\end{equation}
reading the $:$ as 'such that'. The generating functions for these (up to offset) are \begin{equation}
f_1(k) \to ((1-x))^{-1}\\
f_2(k) \to ((1-x)(1-x^2))^{-1}\\
f_3(k) \to ((1-x)(1-x^2)(1-x^3))^{-1}\\
f_3(k) \to ((1-x)(1-x^2)(1-x^3)(1-x^4))^{-1}
\end{equation}
so we can see $f_\infty(k)$ will have the generating function \begin{equation}
f_\infty(k) \to \frac{1}{(x;x)_\infty^2}=\frac{1}{\phi(x)^2}
\end{equation}
with $(a;q)_k$ the q-pochammer symbol and $\phi(x)$ the Euler function. Then $f_\infty(k)$, will give the number of partitions of $k$ into parts of $2$ kinds, A000712.
If we change the equalities to less than symbols \begin{equation}
g_1(k)=\sum_{i_1:i_1<k}i_1 \\
g_2(k)=\sum_{i_1,i_2:i_1+2i_2<k} i_1i_2 \\
g_3(k)=\sum_{i_1,i_2,i_3:i_1+2i_2+3i_3<k} i_1i_2i_3 
\end{equation}
then it would appear that $g_\infty(k)$ gives A000713, with generating function \begin{equation}
g_\infty(k) \to \frac{1}{(1-x)\phi(x)^2}
\end{equation}

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