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 polynomials 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](https://en.wikipedia.org/wiki/Hypertranscendental_function) 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!



Is the function $F(x) = \exp(x) + \exp(\exp(x))x$ a hypertranscendental function?
Yes, $F(x)=e^x + x e^{e^x}$ satisfies an algebraic differential equation, and we can find it explicitly.


Looking at the expressions for $F$ and $F'$, we find that
$$x(F'-e^x)=(1+xe^x)(F-e^x).$$


We can rewrite this equation and its derivative as
\begin{align}xe^{2x}+\quad\quad\quad\ (1-x-xF)e^x &= F-xF'\\
(-1-2x)e^{2x}+(x+F+xF+xF')e^x &= xF''
\end{align}


We can solve these as linear equations for $e^x$ and $e^{2x}$, which give
\begin{align}
e^x&=\frac{x^2(2F'-F'')+x(F'-2F)-F}{x^2(F-F'+1)-x-1}\\
e^{2x}&=\frac{x^2(F'+FF'+F'^2-F''-FF'')+x(F''-F-F^2)-F^2}{x^2(F-F'+1)-x-1}
\end{align}


Using the right hand sides in the equation $(e^x)^2=e^{2x}$ gives an algebraic differential equation for $F$. One way to write it is:


$$(h + 2 h x - 
 x^2 F'')^2 =\\ \Big((1 + F) (x^2-x)-1-h x\Big) \Big(x F''(1 - x - xF)+h(2 F + h + x + xF)
 \Big)$$
where $h=xF'-F$.


