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 = ^\infty a_kx^k then we can describe the inverse term by P^{-1}(x) = ^\infty {na_1^n}\left((-1)^{s+t+u+\cdots}{s!t!u!\cdots}\left({a_1}\right)^s\left({a_1}\right)^t\cdots\right)(x-a_0)^n [Morse and Feshbach (1953)], where, critically s + 2t + 3u + ⋯ = n − 1, all of s, t, u, ⋯ 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 = ^\infty a_kx^k this time define the transformed polynomial as [P](x) = ^\infty \left((sa_1+ta_2+ua_3+\cdots)\right)(x-a_0)^n\\ s+t+u+\cdots=n \\ s\ge t \ge u \ge \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 [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 ak terms. If we use P(x)=x, then we find that the transform is the generating function for A006128. [x] = {^n(1-x^k)} likewise for P(x)=x², we have the generating function for A096541. For P(x)=x + x² + x³ + x⁴ + ⋯, we get the generating function for A066186. For P(x)=x − x² + x³ − x⁴ + ⋯, we get the generating function for A066897. For P(x)=x + 2x² + 3x³ + 4x⁴ + ⋯, we get the generating function for A066184. For P(x)=x + x³ + x⁵ + ⋯, we get A207381. To generate the generating function for the natural numbers we need to put in the function FN(x)=x − x² − x³ − x⁴ − x⁵ + x⁶ − x⁷ + 3x⁸ + 0x⁹ + ⋯, the rule is not clear. Interestingly if we input P(x)=FN(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 FP(x)=2x − 3x² − x³ − x⁵ + x⁶ − 4x⁷ + 3x⁸ + 0x⁹ + ⋯, the rule is not clear. If we input P(x)=FP(x)−FN(x), we do seem to get the generating function for A014689, which makes sense. To get π(n), we can use the coefficients 1, −1, −2, 0, −1, 3, −2, 2, 2, ⋯ CHANGE INEQUALITIES If we force the inequalities to be strict, that is s > t > u > ⋯, then with P(x)=x we have the g.f. of A005895. with P(x)=x + x² + x³ + x⁴ + ⋯, we have the g.f. of A066189. TYPE 2 What about the following transform P(x) = a_0 + a_1 x + a_2 x^2 + \cdots = ^\infty a_kx^k then we can describe the inverse term by T_2[P(x)] = ^\infty \left((-1)^{s+t+u+\cdots}a_0^sa_1^ta_2^ua_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_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 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 = ^\infty x^{4k+1}-x^{4k+2} with P(x)=1 + x², 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 + 3x² + 4x³ + ⋯, then we have T_2\left[{(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[{(x-1)^2}\right]= x {1+mx^m} where the bottom equaility is due to A022693. It would also seem that T_2[1+2x]= {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[{(x-1)^2}\right]= x {1+mx^m}\\ T_2\left[{x-1}\right]={ (1-3x^m)}\\ T_2\left[{1-x}\right]= \eta(t)}{\eta(t^2)}\\ T_2\left[{1+x}\right]= x^{25/24}}{\eta(-\tau)}\\ T_2\left[{x^2-1}\right]=\eta(t^2)}{\eta(t)}\\ T_2\left[{1-x^2}\right]= x^{23/24}\left({16}\right)^{1/24}\\ T_2\left[{1-x^3}\right] = \eta(t)\eta(t^6)}{\eta(t^2)\eta(t^3)}\\ T_2\left[{x^3-1}\right] = \eta(t^3)}{\eta(t)}\\ T_2\left[{x^4-1}\right] = \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] = \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] = \eta(t^4)}{\eta(t^3)}\\ T_2\left[ {x^6-1} \right] = \eta(t^6)}{\eta(t^2)}\\ T_2\left[ {x^6-1} \right] = \eta(t^6)}{\eta(t^3)}\\ t^n={2\pi i}\\ \tau^n={2\pi i}\\ m=\lambda\left(t^2\right) with η(q), Dedekind Eta function, λ, 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)=i_1 \to 1,2,3,4,5,6,7,\cdots\\ f_2(k)= i_1i_2 \to 1,2,5,8,14,20,30,\cdots\\ f_3(k)= i_1i_2i_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∞(k) will have the generating function f_\infty(k) \to {(x;x)_\infty^2}={\phi(x)^2} with (a; q)k the q-pochammer symbol and ϕ(x) the Euler function. Then f∞(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)=i_1 \\ g_2(k)= i_1i_2 \\ g_3(k)= i_1i_2i_3 then it would appear that g∞(k) gives A000713, with generating function g_\infty(k) \to {(1-x)\phi(x)^2}