Consider a general template to generate sequences (or polynomials) using the inverse Mellin transform and a kernel function ϕ(s) $$ p_k(x) = f(x) ^{-1}[\phi(s) q_k(s)](x) $$ here pk(x) and qk(x) are polynomials, and f(x) is a function that cancels out with the generating form from the inverse Mellin transform. This is observed with an example setting qk(s)=sk, ϕ(s)=Γ(s) and f(x)=ex, we have $$ B'_k(x) = e^x ^{-1}[\Gamma(s) s^k](x) $$ where B′k(x) appear to be some form of alternating Bell polynomials, and the coefficients of these polynomials are made up of Stirling numbers of the second kind S₂(n, k) as $$ B'_n(x) = ^n (-1)^{n-k} S_2(n,k) x^k $$ we also find that $$ ^n S_2(n,k)}{2^n} x^{k/2} = e^{} ^{-1}[\Gamma(2s) s^n](x) $$ very interestingly $$ (1+x)^{n+1} ^{-1}[\Gamma(s)\Gamma(1-s) s^n](x) ^n (-1)^{n-k-1} A[n,k] x^{k+1}, k>0 $$ where A(n, k) as the Eulerian numbers. The agreement is off slightly for k = 0. There is a more general form to this $$ (1+x)^{n+t} ^{-1}[{\Gamma(t)} s^n](x) $$ which for t = 1 gives the Eulerian numbers, and for t = 2 is related to A199335. We can even insert t = 1/2, and get a sequence which is related to A185411 (with an additional factor to 1/2n). FIXING THE SIGNS We now consider a modification to the transform to fix the signs, define the inverse-Q transform as $$ p_n(x) = ^{-1}[\phi(s)](n,x) = ^{-1}[\phi(s) (-s)^n](-x) $$ where we have chosen the inverse because of the inverse Mellin transforms, now we have $$ ^{-1}[\Gamma(s)](x) ^{-1}[\Gamma(s)](n,x) = B_n(x) = ^n S_2(n,k) x^k $$ for Bell polynomials Bn(x) and interpreting 0⁰ as 1 which is common in combinatorics. It’s still (perhaps) not entirely right, because for ϕ(s)=Γ(s)Γ(1 − s) we have $$ (1-x)^{n+1}^{-1}[\Gamma(s)\Gamma(1-s)](n,x) = x A_n(x), n>0 $$ relating to Eularian polynomials, equally one could say $$ (1-x)^{-1}[\Gamma(s)\Gamma(1-s)](x) = {(1-x)^n}, n>0 $$ TABLE OF RELATIONS |c|c| Function & Function & Numbers Γ(s) & ex & StirlingS2 Γ(s)Γ(1 − s) & (1 + x)n + 1 & Eulerian Numbers we can see that the function f(x) is clearly related to ℳ−1[ϕ(s)], which is exciting because, by assuming qk(s)=sk for all inputs it links the function ϕ(s) directly a special class of numbers T(n, k). We can as questiosn such as , which kernel ϕ(s) produces the binomials?
In a previous article [1], we found a specific integral transform 𝒬, such that [^{-1}[f(s)](x)](t) = ^{-1}[f(s)](t) for the case that f(s)=Γ(s)ζ(s), as this had a functional equation which could be used to define an invariance. In this work we create a generalised version of this. If we have a general functional relationship for f(x) $$ f(s) = h(s)f(g(s)) $$ where g(s) has a nice inverse g−1(s). Then the prescription for getting a 𝒬 that meets the requirement of equation 1 is finding a kernel to define the transform 𝒬 for an inverse power as [x^{-s}](q,s) = \int_0^\infty x^{-s} k(x,q) \; dx = q^{-g^{-1}(s)} (s))}{g'(g^{-1}(s))} Thus for a test function ϕ(s), $$ [^{-1}[\phi]]=\left[{2\pi i}^{c+i\infty} x^{-s}\phi(s)\;ds \right] $$ $$ [^{-1}[\phi]] = {2\pi i}^{c+i\infty} \left[x^{-s}\right]\phi(s)\;ds $$ $$ [^{-1}[\phi]] = {2\pi i}^{c+i\infty} q^{-g^{-1}(s)} (s))}{g'(g^{-1}(s))}\phi(s)\;ds $$ by letting s → g(t) we get $$ [^{-1}[\phi]] = {2\pi i}(c-i\infty)}^{g^{-1}(c+i\infty)} q^{-t} h(t)\phi(g(t))\;dt = ^{-1}[h(t)\phi(g(t))] $$ the condition for this to remain an inverse Mellin transform (up to sign) is that g−1(c + i∞) → d ± i∞ and g−1(c − i∞) → d ∓ i∞. When ϕ(s)=f(s) which satisfies the functional equation, then equation 1 is satisfied. GETTING THE KERNEL FUNCTION Because we have chosen our transform to be the inverse Mellin transform, we can extract our kernel function by taking the inverse Mellin transform of the desired result of 𝒬[x−s]. Then the kernel function should be given by $$ k(q,x) = ^{-1}\left[ q^{-s} {g'(s)} \right](x) $$ EXAMPLE Take the Riemann zeta function again, but this time without the additional gamma function $$ \zeta(s) = 2^s \pi^{s-1}\sin\left({2}\right) \Gamma(1-s)\zeta(1-s) $$ Here we have $$ h(s) = 2^s \pi^{s-1}\sin\left({2}\right) \Gamma(1-s) \\ g(s) = 1-s \\ g^{-1}(s) = 1-s\\ g'(x) = -1 \\ g^{-1}(c\pm i\infty) = d \mp i \infty $$ we may need to include an additional negative factor due to the integral limits swapping sign. We assemble $$ k_f(q,x) = ^{-1}\left[ -q^{-s} 2^s \pi^{s-1}\sin\left({2}\right) \Gamma(1-s) \right](x) = {q x}\right)}{q x} $$ If we take the modular form $$ y(z) = y\left({z+2}\right) $$ then $$ h(s) = 1\\ g(s) = {s+2} \\ g^{-1}(s) = {s-2} \\ g'(s) = {(s+2)^2} \\ $$ there is a sign swap $$ k_y(q,x) = ^{-1}\left[ q^{-s} (s+2)^2 \right](x) $$ this doesn’t work so well. Starting from a simpler example $$ f(x) = \Gamma(x)f(1-x) $$ gives k(x, q)= − e−qx, then $$ [x^{-s}] = -q^{s-1}\Gamma(1-s) $$ which apparently is connected to the Laplace Transform. Likewise $$ f(s) = (-i)^s\Gamma(s)f(1-s) $$ gives the kernel −e−iqx which will relate to the Fourier transform. For the functional equation $$ f(s) = \Gamma(s)f\left({2}\right) $$ we get $$ k(q,x) = ^{-1}[q^{-s}s^{-1}\Gamma(s)] = \Gamma(0,qx) $$ This defines a transform such that $$ [x^n] = {(n+1)q^{n+1}}, q>0 $$ which seems quite fundamental in terms of differentiation. Another interesting one satisfies $$ f(s) = \Gamma(s)f(s+s^2) $$ which gives $$ k(q,x) = e^{-q x} - + () $$ and $$ f(s) = \Gamma(s)f(s+{2}) $$ $$ k(q,x) = e^{-q x} - qx \Gamma(0,qx) $$ for which $$ [x^n] = {(n+2)q^{n+1}}, q>0 $$ REFERENCES [1] - Riemann Zeta Invariance Under Composed Integral Transform,
From a question I asked online [1], I had deduced that the Laplace transform could be absorbed into the inverse Mellin transform as ^{-1}[\phi] = -^{-1}[\phi^*] and the Mellin transform could be absorbed into the inverse Laplace transform as ^{-1}[\psi] = \Gamma(q)^{-1}[\psi^*] where \phi^* = \Gamma(t)\phi(1-t) and \psi^* = \psi(-e^{-s})e^{-s} the term of ϕ(1 − t) reminded me of the Riemann function equation for ζ(s) which is \zeta(s) = 2^s \pi^{s-1}\sin\left({2}\right) \Gamma(1-s)\zeta(1-s) the question I was then interested in was WHAT OTHER TRANSFORM WHEN APPLIED TO THE INVERSE MELLIN TRANSFORM OF A FUNCTION, WOULD RESULT IN THIS FUNCTIONAL EQUATION, or what is the transform such that ζ(s) is invariant to? The more fundamental quantity in terms of Mellin transforms is Γ(s)ζ(s) which has the integral (Mellin transform) representation: \Gamma(s)\zeta(s) = \int_0^\infty }{e^x-1} \; dx = \left[{e^x-1}\right] we would like to find an integral transform of a function f 𝒬[f] such that [^{-1}[\phi(s)]] = ^{-1}[2^s \pi^{s-1}\sin\left({2}\right)\Gamma(s)\phi(1-s)] such that [^{-1}[\Gamma(s)\zeta(s)]] = ^{-1}[\Gamma(s)\zeta(s)] by virtue of the integral equation 6 we should have something like \left[{e^x-1}\right](s) = {e^s-1} we expect 𝒬 to somewhat resemble a Laplace transform because of the equation ^{-1}[\phi(s)] = ^{-1}[\Gamma(s)\phi(1-s)] Seems that we want something such that [x^{-s}] = \left({2\pi}\right)^{s-1} {\pi} \sin\left({2}(1-s)\right)\Gamma(1-s) the trick seems to be using an inverse Mellin transform on the above to get the relationship Q[x^{-s}] = \int_0^\infty x^{-s}{2 \pi x} \right)}{\pi x}\; dx = q^{-s} 2^s \pi^{s-1} \Gamma(s) \sin\left({2}\right) this is still not quite right as we want to invert the s → 1 − s. It does seem (numerically) for a small region of s values (between 0 and 1?) that [x^{-s}] = \int_0^\infty x^{-s}{2 \pi} \right)}{\pi}\; dx = \left({2\pi}\right)^{s-1} {\pi} \sin\left({2}(1-s)\right)\Gamma(1-s) as required. Hence our transform becomes (note the minus sign) [f] = -\int_0^\infty f(x){2 \pi} \right)}{\pi}\; dx which should (formally) satisfy [^{-1}[\Gamma(s)\zeta(s)] = ^{-1}[\Gamma(s)\zeta(s)] or then ’fixing’ the inverse Mellin transform as given, Γ(s)ζ(s) is some kind of eigen-function of the transform 𝒬... CHECKING THIS FOLLOWS THROUGH Thus $$ [^{-1}[\phi]]=\left[{2\pi i}^{c+i\infty} x^{-s}\phi(s)\;ds \right] $$ $$ [^{-1}[\phi]] = {2\pi i}^{c+i\infty} \left[x^{-s}\right]\phi(s)\;ds $$ $$ [^{-1}[\phi]] = {2\pi i}^{c+i\infty} \left({2\pi}\right)^{s-1} {\pi} \sin\left({2}(1-s)\right)\Gamma(1-s)\phi(s)\;ds $$ by letting s − 1 → −t we get $$ [^{-1}[\phi]] = {2\pi i}^{c'+i\infty} q^{-t} 2^t \pi^{t-1} \sin\left({2}\right)\Gamma(t)\phi(1-t)\;dt = {2\pi i}^{c'+i\infty} q^{-t}\phi^*(t)\;dt = ^{-1}[\phi^*]} $$ where $\phi^*(t) = 2^t \pi^{t-1} \sin\left({2}\right)\Gamma(t)\phi(1-t)$. If we set ϕ(t)=Γ(t)ζ(t) according to the Riemann functional equation we have \zeta(s) = 2^s \pi^{s-1} \sin\left({2}\right) \Gamma(1-s)\zeta(1-s) thus ϕ*(t)=ϕ(t)=Γ(t)ζ(t). CONCLUSION It is formally possible to define such an integral transform. This may be possible and have better convergence for other functional relationships. REFERENCES [1] - APPENDIX If we define the forward transform as _1[f(x)](k) = \int_0^\infty {e^{kx}-1} \;dx we find that _1[x^{s-1}](k) = k^{-s} \Gamma(s)\zeta(s), \; s>1, t>0 or equivalently _1[x^{s-1}](k) = k^{-s} 2^s \pi^{s-1}\sin\left({2}\right)\Gamma(s)\Gamma(1-s)\zeta(1-s), \; s>1, t>0 Thus $$ _1[^{-1}[\phi]]=_1\left[{2\pi i}^{c+i\infty} x^{-s}\phi(s)\;ds \right] $$ $$ _1[^{-1}[\phi]] = {2\pi i}^{c+i\infty} _1\left[x^{-s}\right]\phi(s)\;ds $$ $$ _1[^{-1}[\phi]] = {2\pi i}^{c+i\infty} q^{s-1}\Gamma(1-s)\zeta(1-s)\phi(s)\;ds $$ by letting s − 1 → −t we get $$ _1[^{-1}[\phi]] = {2\pi i}^{c'+i\infty} q^{-t}\Gamma(t)\zeta(t)\phi(1-t)\;dt = {2\pi i}^{c'+i\infty} q^{-t}\phi^*(t)\;dt = ^{-1}[\phi^*]} $$ where ϕ*(t)=Γ(t)ζ(t)ϕ(1 − t). Although this is cool, it’s not quite what we want.
ABSTRACT We consider a ’Fekih-Ahmed’ transform based on a single equation which is probably a coincidence. MAIN Consider the series expansion for the reciprocal Gamma function {\Gamma(z)} = ^\infty a_n z^n = z + \gamma z^2 + \left({2} - {12}\right)z^3 + \cdots According to the Wikipedia page (Reciprocal gamma function) there is an integral formula for these coefficients due to Fekih-Ahmed: a_n = {\pi n!}\int_0^\infty e^{-t}\Im[(\log(t)-i\pi)^n]dt excellent, looks interesting. Let’s draw analogy from the Mellin transform, and the definition of the Gamma function \Gamma(z) = \int_0^\infty e^{-t}t^{z-1} dt both of these equations have an e−t, the integral domain is the same and both relate to Gamma functions. We also note the presence of the term $$ \chi(n) = {n!} $$ which is critical in the Ramanujan master theorem allowing us to express the Mellin transform of a function with expansion $$ f(x) = ^\infty \chi(k)\phi(k)x^k $$ as [f](s) = \int_0^\infty x^{s-1}f(x) \; dx = \Gamma(s)\phi(-s) where possible. Based on this we absorb the coefficient in the definition and define the Fekih-Ahmed transform as [f](n) = \int_0^\infty f(x) \Im[(\log(t)-i\pi)^n] {\pi} and allow functions to be defined by g(x) = ^\infty \chi(k)[f](k)x^k namely {\Gamma(z)} = ^\infty \chi(k)[e^{-x}](k) z^k Put into words: “The gamma function is Mellin transform of e−x, the function whose alternating exponential coefficient is 1.” and “The alternating exponential coefficient of the reciprocal gamma function is the Fekih-Ahmed transform of e−x”. This is not quite symmetric, but there seems to be a rough interplay between the concepts. Let’s explore further... EXAMPLES It seems that $$ {n!}\left[{(x+1)^k}\right](n) = {(k-1)!}, n<k $$ we are in sketchy territory for inserting random functions into the transform, as we are trying to match up terms from series expansions and from the integral evaluations. It will be more productive to take a series expansion of a related function such as $$ {\Gamma(z)\Gamma(z+1)} = z+ 2 \gamma z^2 + \cdots $$ and work out the contents of the FA transform that match this, perhaps it will be a simple function: Following the logic of Mellin transforms, it might be best to explore hypergeometric type arguments. For example the inverse Mellin transform of Γ(s)² is $$ ^{-1}[\Gamma(s)^2](x) = 2K_0(2\sqrt(x)) $$ so we consider $$ g(x) = ^\infty \left[2K_0(2)\right](k+1) z^k = 1 + 2\gamma z + 2\gamma^2 z^2 + {3}z^3 + {3}z^4 + \cdots = A + B + C + \cdots $$ which seems to have a component of $$ A = ^\infty {n!}\gamma^n z^n = e^{2 \gamma z} $$ this seems to be the right kind of language to describe these expansions. We can inspect in more detail the meaning of the term in the integral. In terms of a probability distribution or similar: $$ [f](1)}{1!} = \int_0^\infty f(x) dx $$ the n = 1 term is the normalization. $$ [f](2)}{2!} = \int_0^\infty -\log(x) f(x) dx $$ the n = 2 term is the expectation of the ’negative log’. $$ [f](3)}{3!} = \int_0^\infty (\log(t)^2/2-\zeta(2)) f(x) dx $$ $$ [f](4)}{4!} = \int_0^\infty (\zeta(2)\log(x) - \log(t)^3/6) f(x) dx $$ The coefficients appear to be A109447, $$ ... $$
GENERATING FUNCTION FOR nth COLLATZ ITERATION We can consider the generating function for the Collatz map applied to the positive integers. Define C(n) = n/2 & n \bmod 2 = 0 \\ 3n+1 & n \bmod 2 = 1 and define the mth composition of the function as Cm(n), such that C₀ = n and C₁(n)=C(n) and C₂(n)=C(C(n)). The generating function for positive integers is G_0(x) = {(1-x)^2} for the first iteration we have numbers 4, 1, 10, 2, 16, 3, ... G_1(x) = {(1-x^2)^2}(4+x+2x^2) for the second iteration giving 2, 4, 5, 1, 8, 10, 11, 2, 14, ... we have G_2(x) = {(1-x^4)^2}(2+4x+5x^2+x^3+4x^4+2x^5+x^6) the next iteration is G_3(x) = {(1-x^8)^2} in general this gives G_n(x) = {(1-x^{2^n})^2} for a polynomial of which seems to be order 2n + 1 − 1 these polynomials appear to be related to the current iteration sequence by the following relationship P_n(x) = \left(^{2^n} C_n(k)x^k\right) + \left(^{2^{n+1}-1} (C_n(k)-2 C_n(k-2^n))x^k\right) which we can write as P_n(x) = \left(^{2^{n+1}-1} C_n(k)x^k\right) -2 \left(^{2^{n+1}-1} C_n(k-2^n)x^k\right) We could consider the Cauchy product of this and the simple series {(1-x^{2^n})^2} = 1 + 2 x^{2^n} + 3 x^{2\cdot2^n} + 4 x^{3 \cdot 2^n} + \cdots what does this mean? This means that for any level of iteration, we can describe the coefficient for any number, however large, using the first few function evaluations and a composition. However the expressions rapidly become complicated, with 2n + 1 terms. What conditions would then be required for a coefficient to be 1? For a given iteration this will depend on the number of ways to write a target number t, as the sum of an integer in the range [1, 2n − 1] and any of [0, 2n, 2 ⋅ 2n, 3 ⋅ 2n, ⋯], for one iteration that’s combinations in [1, 2, 3]+[0, 2, 4, 6, 8, ⋯] which can make [1, 2, 3],[3, 4, 5],[5, 6, 7] and so on indicating there are multiple ways to make 3, 5, 7, ⋯. All of the coefficient terms are positive which is nice. The only way a coefficient can be 1 in this iteration is if it is 1 in the polynomial, and multiplied by the 1 in the expanded series. This means we can look at a subset of the polynomial, namely Pn(x). We can then ask, how can a coefficient become 1 in Pn(x)? We can see that Cn(k)>2Cn(k − 2n) for k ∈ [2n + 1, 2n + 1 − 1] to keep the terms positive and non-zero.
If we take the Laurent expansion of the Riemann zeta function about s = 1 $$ \zeta(s) = {s-1} + ^\infty {n!}\gamma_n (s-1)^n $$ which defines γn, the Stieltjes constants, where γ₀ is the Euler-Mascheroni constant. Next perform a series reversion on this to give a series $$ \chi(s) = 1+{s}+^\infty {s^n} $$ which has expansion \chi(s) = 1 + {s} + {s^2} + {s^3} + {2s^4} + \cdots The coefficients κ(n) seem to decrease quite steadily, even up n being a few hundred, where the γn get large. n (n) ---- --------------- -- 0 1.000000000 1 1.000000000 2 0.5772156649 3 0.4059937693 4 0.3135616752 5 0.2556464523 6 0.2159181431 7 0.1869526867 8 0.1648872027 9 0.1475121704 10 0.1334717457 11 0.1218874671 12 0.1121649723 13 0.1038876396 14 0.09675470803 15 0.09054358346 : The first 16 coefficients of the inverse function. Letting $$ R_n=^n i_k $$ and $$ P_n=^n ki_k $$ and {i}n = {i₁, i₂, ⋯|P = n − 1}, I have observed the expression for κ(n) from series reversion to be $$ \kappa(n)=} (-1)^n\left[^{R-1}j-n\right]\left[^n {i_k!}\left({k!}\right)^{i_{k+1}}\right]\gamma_0^{i_1} $$ where we define κ(0)=1. Two examples $$ \kappa(3) = \gamma_0^2 - \gamma_1 = - \left[^{i_1+i_2+i_3-1} (j-n)\right]{1!}\right)^{i_2}\left({2!}\right)^{i_3}}{i_1!i_2!}\gamma_0^{i_1} $$ $$ \kappa(4) = \gamma_0^3 - 3 \gamma_0\gamma_1 + {2} = \left[^{i_1+i_2+i_3+i_4-1} (j-n)\right]{1!}\right)^{i_2}\left({2!}\right)^{i_3}\left({3!}\right)^{i_4}}{i_1!i_2!i_3!}\gamma_0^{i_1} $$ We can conjecture that $$ \kappa(n+1) < \kappa(n), \;\;\; n\in^{>0}? $$ Is this perhaps a more well behaved way to look at the Stieltjes constants?
Well known Dirichlet series include: \zeta(s) = ^\infty {k^s}\\ {\zeta(s)} = ^\infty {k^s}\\ {\zeta(s)} = ^\infty {k^s} If we define the inverse Dirichlet transform to map like \zeta(s) \to 1 \\ {\zeta(s)} \to \mu(k) \\ {\zeta(s)} \to \lambda(k) then we can track less standard variations by creating ratios of zeta functions: {\zeta(s)}\to A210826(k) \\ \zeta(s)^2 \to d(n) = \tau(n) \\ {\zeta(s^2)} \to (k)\\ {\zeta(s^2)} \to {k^q} \\ {2})}{\zeta(s+{2})} \to {k} \\ = ? Consider the duality of ζ(s)→Γ(s) with inverse Mellin transform... DIRICHLET SHIFT OPERATOR We could consider an operator O− (O+) which shifts the argument of a zeta function by −1 (+1). Then use something like the product and quotient rules, to define this on Dirichlet generating functions O^-[\zeta(s)] = \zeta(s-1) \\ O^-[\zeta(s)\zeta(s-1)] = \zeta(s-1)^2 + \zeta(s)\zeta(s-2)\\ O^-[{\zeta(s)}] = {\zeta(s)^2} we can then define the ’number theoretic derivative’ of a function for this latter one we appear to have {\zeta(s)} \to \lambda(n)*A000188(n)\\ {\zeta(s)^2} \to A074722(n) so as ζ(2s)/ζ(s)→λ(n) we could say $$ \delta \lambda(n) = \lambda(n)*A000188(n) - A074722(n) = 0,-1,-2,0,-4,1,-6,-4,-2,1,-10,-4,-12,1,-2,-2,-16,\cdots $$ where clearly, we have every prime, δλ(p)=1 − p, this is nice. We have depending on definition \delta \mu(n) = -A007431(n) \mu(n)-A007431(n) this is $$ \delta \mu(n) = - \phi(d) \mu({d}) $$ and therefore $$ \phi(n) = - \delta \mu(d) $$ then \delta |\mu(n)| = \delta {\zeta(2s)} = {\zeta(2s)^2} = {\zeta(2s)} - {\zeta(2s)^2} \\ \delta |\mu(n)|= A063659(n) - H(n) where H(n)=1 unless n = 9, 16, 18...?, and 2 otherwise. It seems that $$ {\zeta(s)} = ^\infty {(k^2)^s} = ^\infty }(k)\phi()}{k^s} $$ then there is the convolutions $$ {\zeta(2s)^2} = |\mu(n)|*_D }(n)\phi() = |\mu({d})|}(d)\phi() = }(n)\phi() + (1-}(n)) = H(n) $$ we have \delta \tau(n) = 2 \sigma_1(n)\\ \delta \sigma_1(n) = n \tau(n) + \sigma_2(n)\\ \delta \sigma_2(n) = n \sigma_1(n) + \sigma_3(n)\\ \delta \sigma_k(n) = n (n) + (n) noting that σ−1(n)=σ₁(n)/n, also then \delta n\tau(n) = 2 \sigma_1(n) \\ \delta \tau(n) = 2 \sigma_1(n)\\ \delta^2 \tau(n) = 2n \tau(n) + 2 \sigma_2(n) \\ \delta^3 \tau(n) = (4+2n)\sigma_1(n) + 2 \sigma_3(n) By thinking carefully about the linearity of derivatives and implying the δ is a linear operator, then we can easily show that \delta ^k \log(n) = n^k \log(n) for all integer k. In fact in general $$ \delta ^k \log^l(n) = n^k \log^l(n) $$ A stunning result appears to be $$ \delta \Lambda(n) = \log(n^{\phi(n)}) $$ for which the DGF is $$ {\zeta(s)^2} - {\zeta(s)} $$ FURTHER OPERATORS Consider the operator $$ \kappa = [{\zeta(s)} \delta \zeta(s)] $$ such that a factor of ζ(s) is applied, the derivative taken, and then the factor removed. $$ \kappa \zeta(s) = {\zeta(s)} \delta \zeta(s)^2 = {\zeta(s)} 2 \zeta(s)\zeta(s-1) = 2 \zeta(s-1) $$ therefore $$ \kappa 1 = k $$ we also have \kappa \zeta(s-1) \to {\zeta(s)} + \zeta(s-2) \\ \kappa k = P(k) + k^2 where P(k) is A018804(k). We can consider more complicated operators $$ \epsilon = 1 + \delta + \delta^2 + \delta^3 + \cdots $$ then $$ \epsilon \zeta(s) \to \zeta(s) + \zeta(s-1) + \zeta(s-2) + \cdots $$ which implies $$ \epsilon 1 = 1 + k + k^2 + k^3 + \cdots = {1-k} $$ KERNEL GUIDED TRANSFORM For the Lambert transform we have the nicve property that $$ ^\infty a_n {1-x^n} = ^\infty b_n x^n $$ where the relationship is $$ b_n = a_n $$ but we can generalise this by realising that $$ ^\infty a_n f(x^n) = ^\infty b_n x^n $$ where $$ f(x) = ^\infty \kappa(n) x^n $$ generates a new transform, we then know for κ(n)=1, we have the Lambert transform. RATIO OF SIGMA If we let $$ \kappa(n)={\sigma_1(n^2)} $$ then it seems if an bn ζa ζb ------- ------------------ ------------------------------- ------------------------------- μ(n) nϕ(n) 1/ζ(s) ${\zeta(s)}$ ϕ(n) n² ${\zeta(s)}$ ζ(s − 2) μ(n)² A034444(n)=d*(n) ζ(s)/ζ(2s) ζ(s)²/ζ(2s) σ₁(n) ... ζ(s)ζ(s − 1) ζ(s − 2)ζ(s)² n σ₂ ζ(s − 1) ζ(s)ζ(s − 2) If we let $$ \kappa(n)=\phi(n) $$ then it seems if an bn ζa ζb ------ ------------------------- ------------------------------- ----------------------------------- ϕ(n) ∑d|nϕϕ ${\zeta(s)}$ ${\zeta(s)^2}$ μ(n) ∑d|nϕμ ${\zeta(s)}$ ${\zeta(s)^2}$ 1 n ζ(s) ζ(s − 1) n $^n gcd(n,k)$ ζ(s − 1) ζ(s − 1)²/ζ(s) If we let $$ \kappa(n)=\mu(n) $$ then it seems if an bn ζa ζb ------ -------- ------------------------------- --------------------------------- ϕ(n) ∑d|nϕμ ${\zeta(s)}$ ${\zeta(s)^2}$