The function $F(x) = \exp(x) + \exp(\exp(x))x$ plays a role in the formulation of the Lagarias inequality:

$$\sigma(n) \le H_n + \exp(H_n) \log(H_n)$$

If we put $x = \log(H_n)$, then this inequality is equivalent to :

$$\sigma(n) \le F(\log(H_n))$$

I wanted to look at some properties of this function and for this I put 

$$a=x,b=\exp(x),c=\exp(\exp(x))$$

We can compute the derivatives of $F$ recursively through some simple "replacement rules":

$$S:=b+ac$$

$$D(n) = 0 \forall n\in \mathbb{N}$$

$$D(a) = 1$$

$$D(b) = b$$

$$D(c) = bc$$

$$D(M) = D(x) M/x + x D(M/x), \text{ where }$$
$M$ is a monomial in the polynomial $D^{(k)}(S)$ and $x$ is a variable from $a,b,c$ in $M$.

After implementing this in Sagemath:

    var("a,b,c")
    
    R.<a,b,c> = PolynomialRing(QQ,(a,b,c))
    d0 = b+a*c
    
    
    def der(poly):
        if poly == b:
            return b
        if poly == c:
            return b*c
        if poly == a:
            return 1
        mons = poly.monomials()
        coeffs = poly.coefficients()
        s = 0
        for i in range(len(mons)):
            m = mons[i]
            cc = coeffs[i]
            if len(m.variables())==0:
                s+=0
            else:    
                v = m.variables()[0]
                s += cc * ( der(v)*R(m/v)+v*der(R(m/v)))
        return s
    
    for k in range(20):
        #print("$$ \\text{",k,"} ",latex(expand(d0)),"$$")
        print(expand(d0))
        d0 = der(d0)

We get the following poylnomiamls in $a,b,c$ in the form $k, D^{(k)}(S)$:

$$ \text{ 0 }  a c + b $$
$$ \text{ 1 }  a b c + b + c $$
$$ \text{ 2 }  a b^{2} c + a b c + 2 b c + b $$
$$ \text{ 3 }  a b^{3} c + 3 a b^{2} c + a b c + 3 b^{2} c + 3 b c + b $$
$$ \text{ 4 }  a b^{4} c + 6 a b^{3} c + 7 a b^{2} c + 4 b^{3} c + a b c + 12 b^{2} c + 4 b c + b $$
$$ \text{ 5 }  a b^{5} c + 10 a b^{4} c + 25 a b^{3} c + 5 b^{4} c + 15 a b^{2} c + 30 b^{3} c + a b c + 35 b^{2} c + 5 b c + b $$
$$ \text{ 6 }  a b^{6} c + 15 a b^{5} c + 65 a b^{4} c + 6 b^{5} c + 90 a b^{3} c + 60 b^{4} c + 31 a b^{2} c + 150 b^{3} c + a b c + 90 b^{2} c + 6 b c + b $$
$$ \text{ 7 }  a b^{7} c + 21 a b^{6} c + 140 a b^{5} c + 7 b^{6} c + 350 a b^{4} c + 105 b^{5} c + 301 a b^{3} c + 455 b^{4} c + 63 a b^{2} c + 630 b^{3} c + a b c + 217 b^{2} c + 7 b c + b $$
$$ \text{ 8 }  a b^{8} c + 28 a b^{7} c + 266 a b^{6} c + 8 b^{7} c + 1050 a b^{5} c + 168 b^{6} c + 1701 a b^{4} c + 1120 b^{5} c + 966 a b^{3} c + 2800 b^{4} c + 127 a b^{2} c + 2408 b^{3} c + a b c + 504 b^{2} c + 8 b c + b $$
$$ \text{ 9 }  a b^{9} c + 36 a b^{8} c + 462 a b^{7} c + 9 b^{8} c + 2646 a b^{6} c + 252 b^{7} c + 6951 a b^{5} c + 2394 b^{6} c + 7770 a b^{4} c + 9450 b^{5} c + 3025 a b^{3} c + 15309 b^{4} c + 255 a b^{2} c + 8694 b^{3} c + a b c + 1143 b^{2} c + 9 b c + b $$

I tried to come up with a differential equation satisfied by $F(x)$ through using the first derivatives and Groebner bases in Singular, but I could not eliminate $b,c$ from the equations.

**Q1)** So my question is, if it is possible to prove that $F(x)$ is [hypertranscendental][1] or that it satisfies a set of differential equations or one differential equation.

**Q2)** I would also be interested in some further ideas which shed light to the coefficients of these polynomials in $a,b,c$.

Thanks for your help!


  [1]: https://en.wikipedia.org/wiki/Hypertranscendental_function