Exponential versus Factorial Revisited

Karol Gryszka Institute of Mathematics University of the National Education Commission, Krakow Podchora̧żych 2 30-084 Kraków Poland

             karol.gryszka@uken.krakow.pl

Abstract

We study the relationship between iterated factorials, the iterated Euler gamma function, and iterated exponentiation. We demonstrate that both the iterated factorial and the iterated gamma function can be represented as a power tower, where all the terms are identical except for the top term. Additionally, we show that the top terms of this representation form a converging sequence. Finally, we prove that this representation can be reversed.

1 Introduction

Consider a sequence of factorials (entry \underline{A000142} in the On-Line Encylopedia of Integer Sequences (OEIS) [7])

1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, …

and a sequence of iterated exponentiation (tetration) of 2 (entry \underline{A014221}):

1, 2, 4, 16, 65536, 2^{65536}.  (1)

The next term in this sequence has 19729 decimal digits. Now consider a sequence of iterated factorials:

3, 6, 720, 720!, …

with the number 720! having 1747 decimal digits (note that the terms 3, 6, 720 appear in order in the OEIS in four different entries, but not in our context). The last two sequences grow, unsurprisingly, very quickly and have garnered some interest in the past. In particular, the question of how to compare these numbers has received some interest recently. One of the recent and interesting results is that of Velleman [8], who proved, among other things, that the largest number that can be written with five symbols, using positive integers written in decimal notation, exponentiation, and the factorial function, is 9^{9}!!! (where by n!!! we mean ((n!)!)!).

Iterated exponentiation, as illustrated by the example (1)), dates back to Euler [2]. More recently, Knoebel [4] and Anderson [1] also investigated iterated exponentiation, including infinite iterated exponentiation. Shapiro and Shapiro [6] considered iterated exponents reduced (mod k). Velleman [8] studied iterated factorials and exponentiation to determine the most efficient way to generate the largest numbers using factorials and exponentiation. He also showed that one power tower can dominate another, and that power towers tend to dominate iterated factorials. Later, inspired by Velleman’s article, the present author [3] compared iterated exponentiation in many scenarios, including power towers with similar heights and so-called periodic towers.

One of the results obtained in [3] was a form of normalization of power towers. In short, it allows any tetration (power tower with equal terms) to be represented by a power tower whose terms are all equal, except for the top term. Before we state that result, we first introduce the notation used throughout the article.

For given k ≥ 1^{a} and n ≥ 0 we define

F_{n}(k) = \begin{cases}k, & if n = 0, \\ F_{n−1}(k!), & if n ≥ 1. \end{cases}

Note that F_{n}(k)! = F_{n}(k!), so for instance,

F_{2}(3) = (3!)! = 6! = 720, F_{3}(4) = F_{2}(24) = F_{1}(24!) ≈ 10^{10^{25.1611}}.

We also adopt a simplified notation for iterated factorials (not to be confused with other notation that use multiple exclamation marks) in order to avoid cumbersome bracketing:

k \underbrace{! ⋯!}_{×n} = F_{n}(k),

meaning, for example 3!!! = 6!! = 720!. Additionally, we use the following notation for a > 1 and b > 1 (assuming the described operation makes sense in real numbers):

E_{n}(a, b) = \begin{cases}b, & if n = 0; \\ a^{E_{n−1}(a,b)}, & if n > 0, \end{cases}

aNote that if k = 1 or k = 2, then the numbers F_{n}(k) form a constant sequence.

a ↑↑ n = E_{n−1}(a, a),

log_{a}^{(n)} b = \begin{cases}1, & if n = 0; \\ log_{a} b, & if n = 1; \\ log_{a}(log_{a}^{(n−1)} b), & if n > 1. \end{cases}

Let us go back to stating the result from [3].

Theorem 1. Suppose a, b > e^{1/e} and a ≠ b. There exists N ≥ 0, k ∈ ℤ and a converging sequence (c_{n})_{n≥N} such that 1 < c_{n} < b and

a ↑↑ n = E_{n+k}(b, c_{n}).

In this article, we explore a new scenario that has not been considered before: the comparison of iterated factorials. We also consider the continuous version of factorials, namely, the Euler gamma function. This leads to several interesting results, including the representation of the iterated gamma function of fixed number as a power tower with regular behaviour. Surprisingly, the opposite is also true: iterated exponentiation can be represented as iterations of a converging sequence under the gamma function. This in turn seems both interesting and unexpected, as according to Velleman’s results [8] if the power tower just slightly surpasses the iterated factorial, the ratio between further iterations of the respective numbers tends to infinity.

Note that the power tower representation of the iterated factorial or the gamma function (or any other number) is useful because knowing the height of the tower and the top term of the tower is sufficient to compare the two numbers. This, combined with the fact that the mutual comparison between the iterated factorial and the iterated exponentiation seems limited to just [8] motivates us to investigate the topic further.

Recall the following lemmas from [3], which will be useful in the next sections.

Lemma 2. Suppose b > a > e^{1/e} and define a sequence

a_{n} = log_{a}^{(n)}(b ↑↑ n).

Then (a_{n})_{n∈ℕ} is an increasing sequence and lim_{n→+∞} a_{n} < +∞.

Lemma 3. Suppose a, b > e^{1/e} and n > 0. Then there is p ≥ 0 and c ∈ (1, b] such that

a ↑↑ n = E_{p}(b, c).

2 Comparing factorials

In this section, we focus on comparing iterated factorials in two ways. First, we directly compare one iterated factorial with another. Then we show that F_{n}(k) can be replaced by an appropriate power tower.

Theorem 4. Suppose n, n′ ≥ 3 and k > k′ are given. If F_{k−k′}(n) > n′, then F_{k}(n) > F_{k′}(n′).

Proof. We have F_{k}(n) = F_{k′}(F_{k−k′}(n)) > F_{k′}(n′).

The above result is not surprising, as we are merely comparing a few initial iterations of the factorial.

The following theorem shows that for each iterated factorial, one can find a “representation” as a power tower. This mimics the result obtained in Theorem 1. In the article, when the base of the logarithm is not given, we mean base 10 logarithms.

Theorem 5. For any k ≥ 3 there is ℓ ∈ ℤ, N > 0 and the sequence (c_{n})_{n∈ℕ} such that for each n ≥ N we have 10 ≥ c_{n} > 1 and

F_{n}(k) = E_{n+ℓ}(10, c_{n}).

The sequence (c_{n})_{n≥N} is increasing and convergent. Moreover, we have

lim_{n→+∞} \frac{E_{n+ℓ}(10, c)}{F_{n}(k) ln F_{n}(k)} = \frac{1}{ln 10}.  (2)

Proof. Note that if n ≥ 2, then F_{n}(k) > 10. Set N_{1} = 2 and for n ≥ N_{1} define

b_{n} := log^{(n−2)} F_{n}(k) = log^{(n−2)} F_{n−2}(F_{2}(k)).  (3)

Then by the result of Velleman [8] there is N_{2} > 0 such that for each n ≥ N_{2} we have

F_{n−2}(F_{2}(k)) < E_{n−2}(F_{2}(k), F_{2}(k)) = F_{2}(k) ↑↑ (n − 1).  (4)

Then for n ≥ 3,

log b_{n} = log (log^{(n−2)} F_{n−2}(F_{2}(k)))
< log (log^{(n−2)} F_{2}(k) ↑↑ (n − 1))
= log^{(n−1)} F_{2}(k) ↑↑ (n − 1).

and by Lemma 2 the sequence of numbers a_{n} = log^{(n−1)} F_{2}(k) ↑↑ (n − 1) is convergent. To prove that (b_{n})_{n≥0} is convergent, it is therefore enough to show that (b_{n}) is increasing.

Notice that for m ≥ 28 we have log m! > m. Indeed, since the latter is equivalent to m! > 10^{m}, using the the well-known bound m! > (m/e)^{m} we have for m ≥ 28:

m! > (\frac{m}{e})^{m} > 10^{m}.  (5)

Furthermore, for m ≥ 3 we have m!! ≥ 3!! = 720, which by (5) implies that

log log log m!!! > log log m!! for any m ≥ 3. In particular, for n ≥ 5, we have

b_{n} = log^{(n−5)}(log log log F_{n−5}(F_{2}(k))!!!)
> log^{(n−5)}(log log F_{n−5}(F_{2}(k))!!)
= log^{(n−3)} F_{n−3}(F_{2}(k)) = b_{n−1},

so (b_{n})_{n≥5} is an increasing sequence. Furthermore,

b_{5} = log^{(3)} F_{3}(F_{2}(k)) ≥ log^{(3)} F_{5}(3) > 1749,

and thus there is b > 1749 such that b_{n} → b.

By Lemma 3 there is p ≥ 0 and c ∈ (1, 10] such that b = E_{p}(10, c). Set b_{n} = E_{p}(10, c_{n}) for suitable choice of c_{n}. Since c_{n} → c ∈ (1, 10] and c_{n} is increasing, there is N ≥ max{5, N_{2}} such that c_{N} > 1. It remains to notice that for n ≥ N, using the definition (3), we have

F_{n}(k) = E_{n−2}(10, E_{p}(10, c_{n})) = E_{n+p−2}(10, c_{n}).

We now show that (2) holds. The argument used in this proof is similar to the one used by Velleman [8]. First, we show that for n ≥ N we have

\frac{F_{n}(k)(ln F_{n}(k) − 1)}{ln 10} ≤ E_{n+ℓ}(10, c) ≤ \frac{F_{n}(k) ln F_{n}(k)}{ln 10}.  (6)

Suppose that for some n ≥ N we have

E_{n+ℓ}(10, c) < \frac{F_{n}(k)(ln F_{n}(k) − 1)}{ln 10}.

Then

E_{n+ℓ+1}(10, c) = e^{E_{n+ℓ}(10,c) ln 10} < e^{F_{n}(k)(ln F_{n}(k)−1)} = (\frac{F_{n}(k)}{e})^{F_{n}(k)} < F_{n}(k)! = F_{n+1}(k).

As c_{n+1} ≤ c, we now have

F_{n+1}(k) = E_{n+ℓ+1}(10, c_{n+1}) ≤ E_{n+ℓ+1}(10, c) < F_{n+1}(k),

which is a contradiction. Next, suppose that for some n ≥ N we have

E_{n+ℓ}(10, c) > \frac{F_{n}(k) ln F_{n}(k)}{ln 10}.

Since (c_{m})_{n≥N} is increasing and the function x ↦ E_{n+ℓ}(10, x) is continuous, there is M ≥ n such that for each m ≥ M we have

E_{n+ℓ}(10, c_{m}) > \frac{F_{n}(k) ln F_{n}(k)}{ln 10} and F_{m}(k) ≥ 10.

Then, using the fact that r^{r} > r! ln r! (r ≥ 3 is a positive integer; see [8] for the proof of that inequality), we get

E_{n+ℓ+1}(10, c_{m}) = e^{E_{n+ℓ}(10,c_{m}) ln 10} > e^{F_{n}(k) ln F_{n}(k)} = (F_{n}(k))^{F_{n}(k)}

> F_{n}(k)! ln F_{n}(k)! = F_{n+1}(k) ln F_{n+1}(k)

> \frac{F_{n+1}(k) ln F_{n+1}(k)}{ln 10}.

Thus, by induction,

E_{n+ℓ+s}(10, c_{m}) > \frac{F_{n+s}(k) ln F_{n+s}(k)}{ln 10}

for all s ≥ 0. This implies that, as m ≥ n, we also have

F_{m}(k) = E_{m+ℓ}(10, c_{m}) > \frac{F_{m}(k) ln F_{m}(k)}{ln 10} ≥ F_{m}(k),

which is another contradiction. Summing up, we have established that (6) hold for all n large enough. But that bounds can be rewritten to

\frac{1}{ln 10} − \frac{1}{ln 10 ln F_{n}(k)} ≤ \frac{E_{n}(10, c)}{F_{n}(k) ln F_{n}(k)} ≤ \frac{1}{ln 10},

from which (2) follows. This concludes the proof.

Remark 6. It is clear that the decimal base in Theorem 5 is not special: the same result can be proved for any base b > e^{1/e} instead of 10. The only significant difference is the estimation for m such that log_{b} m! > m, or m/e > b.

Example 7. Let k = 9, then (with a little help from WolframAlpha [9]), we get

9! = 362 880,
9!! = 10^{10^{6.269497348270999...}} ≈ E_{2}(10, 6.269497348270999),
9!!! = 10^{10^{10^{6.269498812196401...}}} ≈ E_{3}(10, 6.269498812196401),
9!!!! ≈ E_{4}(10, 6.269498812196401).

This suggests that

F_{n}(9) ≈ E_{n}(10, 6.269498812196401)

for all n ≥ 3. This approximation is correct and justified by Theorem 5. Furthermore, a stronger approximation can be provided:

\frac{E_{n}(10, 6.269498812196401)}{F_{n}(9) ln F_{n}(9)} ≈ 1.

We can easily find out that the following approximation is also correct:

336 500 ↑↑ n ≈ E_{n}(10, 6.2694) ≈ F_{n}(9).

Thus it is interesting to formalize the approximation. We present the result below.

Theorem 8. For any k ≥ 3 there is b_{n} > e^{1/e} such that F_{n}(k) = b_{n} ↑↑ n. The sequence (b_{n})_{n∈ℕ} converges to some b ≤ k!.

Proof. Write a_{n} := F_{n}(k). It follows from a standard intermediate-value argument that one can find b_{n} such that b_{n} ↑↑ n = F_{n}(k) (for similar reasoning we refer to [3]). We are thus left to show that (b_{n})_{n∈ℕ} is bounded and eventually increasing.

Using an argument similar to (3) and (4) we get for sufficiently large n that

F_{n}(⌊b_{n+1}⌋) < b_{n+1} ↑↑ (n + 1) = F_{n}(k!)

and thus b_{n} ≤ k! for all n large enough.

We now show that b_{n} < b_{n+1} for all n large enough. First, we show that for sufficiently large n we have

\frac{b_{n} ↑↑ n}{e} > b_{n}.  (7)

This is equivalent to

b_{n}^{b_{n}↑↑(n−1)−1} > e,

or to

b_{n} ↑↑ (n − 1) > log_{b_{n}} e + 1.  (8)

Recall that if s > e^{1/e}, then s ↑↑ n → +∞ and if e^{−1/e} ≤ s ≤ e^{1/e}, then s ↑↑ n converges (see [1]). As F_{n}(k) diverges for k ≥ 3, from that we get b_{n} > e^{1/e} and log_{b_{n}} e < e, which implies (8) and thus (7) is established for all sufficiently large n.

Finally, we write

b_{n+1} ↑↑ (n + 1) = (b_{n} ↑↑ n)! > (\frac{b_{n} ↑↑ n}{e})^{b_{n}↑↑n} > b_{n}^{b_{n}↑↑n} = b_{n} ↑↑ (n + 1),

conclude the inequality b_{n} < b_{n+1}, and finish the proof.

3 The case of Euler gamma function

We now move our attention to the continuous version of the factorial. Here, the Euler gamma function is considered on positive real numbers

Γ(x) := ∫_{0}^{+∞} t^{x−1}e^{−x}dt.

\begin{matrix}1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ x \\ y \end{matrix}

Figure 1: Solutions to Γ(x) = x.

The equation Γ(x) = x has two positive solutions (see Figure 1). There is one trivial solution Γ(1) = 1, the second one is z ≈ 3.562. If we let Γ^{n}(x) denote the n-th iteration of x under Γ (i.e., Γ^{0}(x) = x, Γ^{n}(x) = Γ(Γ^{n−1}(x)) for n > 0), then for each x > z we have lim_{n→+∞} Γ^{n}(x) = +∞. Thus, from now on we consider the case x > z.

Our goal is to extend Theorem 5 to the gamma function. Note that since Γ(x) = (x − 1)! for x ∈ ℕ \ {0}, the extension improves the theorem to real values above x = 3.

We also highlight to the reader that throughout the rest of the paper, the constant “z” is reserved for the non-trivial solution of Γ(x) = x.

We now establish a generalization of Theorem 5 to the gamma function. We begin with the following lemma, which provides a weaker (yet more useful for the purpose of this article) version of the inequality obtained in [5], but with a proof that is considerably more elementary.

Lemma 9. For all x ≥ 1 we have

Γ(x + 1) > (\frac{x}{e})^{x} .  (9)

For all x > 4 we have

Γ(x + 1) ln Γ(x + 1) ≤ Γ(x)^{Γ(x)}.  (10)

Proof. First, we show that the inequality (9) holds for x ∈ [1, 2]. For that, let us define the function f(x) = ln (\frac{x}{e})^{x}. Then for x ∈ [1, 2] we have f(x) < 0. On the other hand, ln Γ(x + 1) ≥ ln Γ(2) ≥ 0 (see Figure 2). This implies that inequality (9) is established for x ∈ [1, 2].

Notice that since (1 − \frac{1}{x})^{x−1} > \frac{1}{e}, we have

x · \frac{(x − 1)^{x−1}}{e^{x−1}} > \frac{x^{x}}{e^{x}} .

\begin{matrix}−1 \\ 1 \\ 2 \\ 3 \\ 4 \\ −1 \\ 1 \\ 2 \\ 3 \\ ln Γ(x + 1) \\ f(x) \\ x \\ y \end{matrix}

Figure 2: Graph of functions ln Γ(x + 1) and f(x) for x > 0.

This means that if the inequality is established for x ∈ [n − 1, n] with integer n ≥ 2, it is also valid in the interval [n, n + 1], since

Γ(x + 1) = xΓ(x) > x · \frac{(x − 1)^{x−1}}{e^{x−1}} > \frac{x^{x}}{e^{x}} .

This remark and simple induction completes the proof of (9).

The inequality (10) is equivalent to

ln(Γ(x + 1))^{x} ≤ Γ(x)^{Γ(x)−1}.  (11)

We apply a simple bound Γ(y) ≤ y^{y−1} (valid for y ≥ 1) to have

(Γ(x + 1))^{x} ≤ ((x + 1)^{x})^{x} = e^{x^{2} ln(x+1)} ≤ e^{x^{3}}

and we apply simple bounds Γ(x) > (3/2)^{x}, Γ(x) > x, both valid for x > 4, to obtain

(Γ(x))^{Γ(x)−1} > ((\frac{3}{2})^{x})^{x−1} = (\frac{3}{2})^{x^{2}−x} .

To show that (11) holds, it is now sufficient to show that for x > 4 we have

x^{3} < (\frac{3}{2})^{x^{2}−x} ,

which is easily verifiable.

Theorem 10. For any b > e^{1/e} and any x ≥ z there is ℓ ∈ ℤ, N > 0 and a sequence (c_{n})_{n∈ℕ} such that for n ≥ N we have 10 ≥ c_{n} > 1 and

Γ^{n}(x) = E_{n+ℓ}(b, c_{n}).

The sequence (c_{n})_{n≥N} is increasing and convergent. Finally, we have

lim_{n→+∞} \frac{E_{n+ℓ}(b, c)}{((Γ^{n}(k) − 1) ln(Γ^{n}(k) − 1)} = \frac{1}{ln b}.  (12)

Proof. Let Y > 0 be such that for all y ≥ Y we have log_{b} Γ(y) > y. The number Y exists because we have, using (9), the following chain of inequalities

log_{b} Γ(y) > log_{b} (\frac{y − 1}{e})^{y−1} > y,

while the last one is true as it is equivalent to

(\frac{y − 1}{e})^{y−1} > b^{y},

which clearly holds for all y ≥ Y with Y ≥ b sufficiently large.

Let N > 0 be such that Γ^{N}(x) > Y and set

b_{n} = log_{b}^{(n)} Γ^{N+n}(x).  (13)

This sequence is well-defined and increasing. Indeed, notice that b_{0} = Γ^{N}(x) and if b_{n} is well-defined for some n ≥ 0, then since we can apply log_{b}( ·) n times to Γ^{N+n}(x) and

log_{b} Γ(Γ^{N+n}(x)) > Γ^{N+n}(x),  (14)

we can apply log_{b}( ·) n times to log_{b} Γ(Γ^{N+n}(x)) so that b_{n+1} is well-defined. We also notice that (14) implies b_{n+1} > b_{n}.

Set k = ⌈x⌉. Then k ≥ 4 and

Γ(x) ≤ Γ(k) = (k − 1)!.

Furthermore,

Γ^{2}(x) < Γ^{2}(k) = Γ((k − 1)!) = ((k − 1)! − 1)! < k!!

and by simple induction,

Γ^{n}(x) < F_{n}(k)  (15)

for all n ≥ 1 (for n = 0 we can have equality).

Consider the sequence (b_{n})_{n≥0} again and notice that by (15),

b_{n} = log_{b}^{(n)} Γ^{N+n}(x) < log_{b}^{(n)} F_{N+n}(k) = log_{b}^{(n)} F_{n}(F_{N}(k))

for all n ≥ 1. Finally, we have F_{N}(k) > Γ^{N}(x) ≥ Y, thus, by the proof of Theorem 5, the sequence d_{n} = log_{b}^{(n)} F_{n}(F_{N}(k)) is convergent.

We have constructed two sequences of numbers (b_{n})_{n≥0} and (d_{n})_{n≥0} such that

• (b_{n})_{n≥0} is increasing,

• b_{n} < d_{n} for n ≥ 1,

• (d_{n})_{n≥0} is convergent.

Thus (b_{n})_{n≥0} is convergent. By (13),

Γ^{N+n}(x) = E_{n}(b, b_{n}).

Let c_{n} ∈ (1, b] and p_{n} be the numbers defined via relation (see Lemma 3)

b_{n} = E_{p_{n}}(b, c_{n}).

Since (b_{n})_{n≥0} is convergent, so is (c_{n})_{n≥0} and the sequence (p_{n})_{n≥0} is eventually constant. More precisely, let c = lim_{n→+∞} c_{n} ∈ (1, b] and p = lim_{n→+∞} p_{n}. Then there is N_{1} ≥ N such that p_{n} = p and c_{n} ∈ (1, 10] for each n ≥ N_{1}. Then for such n,

Γ^{N+n}(x) = E_{n+p}(b, c_{n})

and finally,

Γ^{n}(x) = E_{n+p−N}(b, c_{n}).

Now we show that (12) holds. For that, the argument is similar to the one presented in the proof of Theorem 5. The goal is to show that

\frac{(Γ^{n}(k) − 1)(ln(Γ^{n}(k) − 1) − 1)}{ln b} ≤ E_{n+ℓ}(b, c) ≤ \frac{(Γ^{n}(k) − 1) ln(Γ^{n}(k) − 1)}{ln b}.  (16)

We apply inequality (9) to show the lower bounds of (16) and we apply inequality (10) for the upper bounds. We omit the details.

Let us perform several calculation of the following form:

10 ≈ Γ(4.39007), 10^{10} ≈ Γ^{2}(4.64095), 10^{10^{10}} ≈ Γ(1160000000) ≈ Γ^{3}(4.5982).

We observe that the arguments of the iterated gamma function are “not that high”, which may suggest that one can replace power towers with the iterated gamma function. This turns out to be true: the process described by Theorem 10 can be reversed. Before addressing this problem, let us prove the following auxiliary lemma.

Lemma 11. For any n > 0 and any b ≥ e we have

Γ(E_{n}(b, b)) > E_{n+1}(b, b).

Proof. We apply the inequality (9) in the following form

Γ(y) > (\frac{y − 1}{e})^{y−1}

again with y = E_{n}(b, b). Then, using b ≥ e and the Bernoulli inequality,

Γ(E_{n}(b, b)) > (\frac{E_{n}(b, b) − 1}{e})^{E_{n}(b,b)−1}

≥ (\frac{E_{n}(b, b) − 1}{b})^{E_{n}(b,b)−1}

= (\frac{E_{n}(b, b)}{b})^{E_{n}(b,b)−1} (1 − \frac{1}{E_{n}(b, b)})^{E_{n}(b,b)−1}

≥ (\frac{E_{n}(b, b)}{b})^{E_{n}(b,b)−1} (1 − \frac{E_{n}(b, b) − 1}{E_{n}(b, b)})

= (\frac{E_{n}(b, b)}{b})^{E_{n}(b,b)−1} · \frac{1}{E_{n}(b, b)}

= (b^{E_{n−1}(b,b)−1})^{E_{n}(b,b)−1} · b^{−E_{n−1}(b,b)}

= b^{(E_{n−1}(b,b)−1)(E_{n}(b,b)−1)−E_{n−1}(b,b)}

= b^{E_{n−1}(b,b)E_{n}(b,b)−E_{n}(b,b)−2E_{n−1}(b,b)+1}.

To finish the proof it is enough to show that

E_{n−1}(b, b)E_{n}(b, b) − E_{n}(b, b) − 2E_{n−1}(b, b) + 1 > E_{n}(b, b),

or, equivalently,

E_{n−1}(b, b) − \frac{2E_{n−1}(b, b) − 1}{E_{n}(b, b)} > 2.

Notice that E_{n−1}(b, b) ≥ e for each n > 0, thus it is enough to show that

\frac{2E_{n−1}(b, b) − 1}{E_{n}(b, b)} < e − 2.  (17)

Let x = E_{n−1}(b, b). Then the inequality can be seen as

\frac{2x − 1}{b^{x}} < e − 2 or 2x − 1 < (e − 2)b^{x}.

Notice again that b ≥ e. Hence

(e − 2)b^{x} ≥ (e − 2)e^{x} > \frac{e^{x}}{2} > 2x − 1,

which implies (17) and completes the proof.

We note that this lemma actually shows the generalization to arbitrary base b > z of the problem investigated by Velemann [8], who solved a simplified version of the above for factorials and the case b = 10.

We are now ready to prove our final result.

Theorem 12. Let b > z. There is a converging (and decreasing for sufficiently large n) sequence (x_{n})_{n≥0} such that b ↑↑ n = Γ^{n}(x_{n}).

Proof. Fix x > z. By Theorem 10 we can find ℓ and N > 0 such that for n ≥ N,

Γ^{n}(x) = E_{n+ℓ}(b, c_{n}).

Here, b ≥ c_{n} > 1 and c_{n} → c. After suitable substitution x ↦ Γ^{ℓ}(x) and correction of N (this substitution preserves the assumption that x > z) we can assume that ℓ = 0, i.e.,

Γ^{n}(x) = E_{n}(b, c_{n}).

In case ℓ < 0 we are interpreting Γ^{−1}(x) to be the unique x′ > z such that Γ(x′) = x. Notice that

Γ^{n}(Γ(x)) = Γ^{n+1}(x) = E_{n+1}(b, c_{n+1}) > E_{n+1}(b, 1) = E_{n}(b, b) ≥ E_{n}(b, c_{n}) = Γ^{n}(x),

hence by the intermediate value theorem for each suitable n there is x_{n} ∈ [x, Γ(x)) such that

Γ^{n}(x_{n}) = E_{n}(b, b).

Let x′_{n} = Γ^{−1}(x_{n}). Thus

Γ^{n+1}(x′_{n}) = E_{n}(b, b).

Shifting indices (i.e., we rename x′_{n} to x′_{n+1}) and simplifying notation x′_{n} → x_{n}, we have

Γ^{n}(x_{n}) = E_{n−1}(b, b) = b ↑↑ n.

Notice that the numbers x_{n} form a bounded sequence and thus to finish the proof it is enough to show that this sequence is monotone.

We have by Lemma 11,

Γ^{n+1}(x_{n}) = Γ(Γ^{n}(x_{n})) = Γ(E_{n−1}(b, b)) > E_{n}(b, b) = Γ^{n+1}(x_{n+1}),

thus x_{n} > x_{n+1} and hence (x_{n})_{n≥0} is a decreasing sequence. This completes the proof.

4 Acknowledgment

The author would like to thank the referee for carefully reading the paper and for providing many valuable comments and suggestions, which significantly improved the article.

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2020 Mathematics Subject Classification: Primary 11B65; Secondary 33B15, 11A67, 26A18. Keywords: gamma function, tetration, power tower, iteration of classic functions, large number representation.

(Concerned with sequences \underline{A000142} and \underline{A014221}.)

Received March 3 2025; revised versions received March 4 2025; April 3 2025. Published in Journal of Integer Sequences, April 5 2025.

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