# Further Sequence Transforms, Sum over infinite set of indices conforming to a rule

## Abstract

I consider defining transforms to mimic the inverse series transform in structure

## Main

If we have an arbitrary series expansion,

$$P(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots=\sum_{k=0}^{\infty}a_{k}x^{k}\\$$

then we can describe the inverse term by

$$P^{-1}(x)=\sum_{n=1}^{\infty}\frac{1}{na_{1}^{n}}\left(\sum_{s,t,u,\cdots}(-1)^{s+t+u+\cdots}\frac{n(n+1)\cdots(n-1+s+t+u\cdots)}{s!t!u!\cdots}\left(\frac{a_{2}}{a_{1}}\right)^{s}\left(\frac{a_{3}}{a_{1}}\right)^{t}\cdots\right)(x-a_{0})^{n}\\$$

[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?

## Example 1

Again let

$$P(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots=\sum_{k=0}^{\infty}a_{k}x^{k}\\$$

this time define the transformed polynomial as

$$\mathcal{T_{1}}[P](x)=\sum_{n=1}^{\infty}\left(\sum_{s,t,u,\cdots}(sa_{1}+ta_{2}+ua_{3}+\cdots)\right)(x-a_{0})^{n}\\ s+t+u+\cdots=n\\ s\geq t\geq u\geq\cdots\\$$

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

$$\mathcal{T_{1}}[P](x)=(a_{1})(x-a_{0})+(3a_{1}+a_{2})(x-a_{0})^{2}+(6a_{1}+2a_{2}+a_{3})(x-a_{0})^{3}+(12a_{1}+5a_{2}+2a_{3}+a_{4})(x-a_{0})^{4}+\cdots\\$$

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.

$$\mathcal{T_{1}}[x]=\sum_{n>0}\frac{nx^{n}}{\prod_{k=1}^{n}(1-x^{k})}\\$$

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 $$1,-1,-2,0,-1,3,-2,2,2,\cdots$$

## 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.

## Type 2

$$P(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots=\sum_{k=0}^{\infty}a_{k}x^{k}\\$$

then we can describe the inverse term by

$$T_{2}[P(x)]=\sum_{n=1}^{\infty}\left(\sum_{s,t,u,\cdots}(-1)^{s+t+u+\cdots}a_{0}^{s}a_{1}^{t}a_{2}^{u}a_{3}^{w}\cdots\right)x^{n}\\ s+2t+3u+\cdots=n-1\\$$

Does this generate anything coherent? We get

$$T_{2}[P(x)]=x-a_{0}x^{2}+(a_{0}^{2}-a_{1})x^{3}+(a_{0}a_{1}-a_{0}^{3}-a_{2})x^{4}+(a_{0}a_{2}+a_{1}^{2}-a_{3}-a_{0}^{2}a_{1}+a_{0}^{4})x^{5}+\cdots\\$$

with $$P(x)=1$$, we will then get

$$T_{2}[1]=x-x^{2}+x^{3}-x^{4}+x^{5}+\cdots\\$$

with $$P(x)=1+x$$, we will then get

$$T_{2}[1+x]=x-x^{2}+x^{5}-x^{6}+\cdots=\sum_{k=0}^{\infty}x^{4k+1}-x^{4k+2}\\$$

with $$P(x)=1+x^{2}$$, we will then get

$$T_{2}[1+x]=x-x^{2}+x^{3}-2x^{4}+2x^{5}-2x^{6}+3x^{7}-3x^{8}+3x^{9}-4x^{10}+\cdots\\$$

nicely if we input $$P(x)=1+2x+3x^{2}+4x^{3}+\cdots$$, then we have

$$T_{2}\left[\frac{1}{(x-1)^{2}}\right]=x-x^{2}-x^{3}-2x^{4}+2x^{5}-x^{6}+4x^{7}-x^{8}+18x^{9}-22x^{10}+\cdots\\ T_{2}\left[\frac{1}{(x-1)^{2}}\right]=x\prod_{m>0}\frac{1}{1+mx^{m}}\\$$

where the bottom equaility is due to A022693. It would also seem that

$$T_{2}[1+2x]=\frac{x}{1+x+2x^{2}+2x^{3}}\\$$

for

$$T_{2}[1-x+x^{2}-x^{3}+x^{4}-\cdots]\\$$

we have the alternating signs of the partition numbers, A000041.

## Possible Results

Below I list a few significant transformations of simple input generating functions. There appears to be a strong connection to modular forms.

$$T_{2}\left[\frac{1}{(x-1)^{2}}\right]=x\prod_{m>0}\frac{1}{1+mx^{m}}\\ T_{2}\left[\frac{3}{x-1}\right]=\frac{x}{\prod_{m>0}(1-3x^{m})}\\ T_{2}\left[\frac{1}{1-x}\right]=\frac{x^{25/24}\eta(t)}{\eta(t^{2})}\\ T_{2}\left[\frac{1}{1+x}\right]=\frac{(-1)^{1/24}x^{25/24}}{\eta(-\tau)}\\ T_{2}\left[\frac{1}{x^{2}-1}\right]=\frac{x^{23/24}\eta(t^{2})}{\eta(t)}\\ T_{2}\left[\frac{1}{1-x^{2}}\right]=x^{23/24}\left(\frac{m(1-m)}{16}\right)^{1/24}\\ T_{2}\left[\frac{1+x}{1-x^{3}}\right]=\frac{x^{11/12}\eta(t)\eta(t^{6})}{\eta(t^{2})\eta(t^{3})}\\ T_{2}\left[\frac{1+x}{x^{3}-1}\right]=\frac{x^{11/12}\eta(t^{3})}{\eta(t)}\\ T_{2}\left[\frac{x}{x^{4}-1}\right]=\frac{x^{11/12}\eta(t^{4})}{\eta(t^{2})}\\ T_{2}\left[1+x+x^{4}-x^{5}+x^{6}-x^{7}+x^{9}+x^{10}-x^{11}+x^{12}+x^{13}-x^{15}+x^{16}-x^{17}+\cdots\right]=\frac{x^{5/6}\eta(t)\eta(t^{12})}{\eta(t^{2})\eta(t^{3})\eta(t^{4})}\\ T_{2}[00-110-102-100000-140-1]=\frac{x^{23/24}\eta(t^{4})}{\eta(t^{3})}\\ T_{2}\left[\frac{x+x^{3}}{x^{6}-1}\right]=\frac{x^{5/6}\eta(t^{6})}{\eta(t^{2})}\\ T_{2}\left[\frac{x^{2}}{x^{6}-1}\right]=\frac{x^{7/8}\eta(t^{6})}{\eta(t^{3})}\\ t^{n}=\frac{n\log(x)}{2\pi i}\\ \tau^{n}=\frac{n\log(-x)}{2\pi i}\\ m=\lambda\left(t^{2}\right)\\$$

with $$\eta(q)$$, Dedekind Eta function, $$\lambda$$, is the modular lambda function.

## How Does This Come About

We can observe a similar set of sums, with restricted tuples of indices.

$$f_{1}(k)=\sum_{i_{1}:i_{1}=k}i_{1}\to 1,2,3,4,5,6,7,\cdots\\ f_{2}(k)=\sum_{i_{1},i_{2}:i_{1}+2i_{2}=k}i_{1}i_{2}\to 1,2,5,8,14,20,30,\cdots\\ f_{3}(k)=\sum_{i_{1},i_{2},i_{3}:i_{1}+2i_{2}+3i_{3}=k}i_{1}i_{2}i_{3}\to 1,2,5,10,18,30,49,\cdots\\$$

reading the $$:$$ as ’such that’. The generating functions for these (up to offset) are

$$f_{1}(k)\to((1-x))^{-1}\\ f_{2}(k)\to((1-x)(1-x^{2}))^{-1}\\ f_{3}(k)\to((1-x)(1-x^{2})(1-x^{3}))^{-1}\\ f_{3}(k)\to((1-x)(1-x^{2})(1-x^{3})(1-x^{4}))^{-1}\\$$

so we can see $$f_{\infty}(k)$$ will have the generating function

$$f_{\infty}(k)\to\frac{1}{(x;x)_{\infty}^{2}}=\frac{1}{\phi(x)^{2}}\\$$

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

$$g_{1}(k)=\sum_{i_{1}:i_{1}<k}i_{1}\\ g_{2}(k)=\sum_{i_{1},i_{2}:i_{1}+2i_{2}<k}i_{1}i_{2}\\ g_{3}(k)=\sum_{i_{1},i_{2},i_{3}:i_{1}+2i_{2}+3i_{3}<k}i_{1}i_{2}i_{3}\\$$

then it would appear that $$g_{\infty}(k)$$ gives A000713, with generating function

$$g_{\infty}(k)\to\frac{1}{(1-x)\phi(x)^{2}}\\$$