On a Sequence Related to the Factoradic Representation of an Integer

Maximiliano Sánchez Garza Facultad de Ciencias Físico Matemáticas Universidad Autónoma de Nuevo León Av. Universidad Ciudad Universitaria San Nicolás de los Garza, Nuevo León 66451 México

               maxsanchez1312@gmail.com

Enrique Treviño Mathematics Department Lake Forest College 555 N. Sheridan Road Lake Forest, IL 60045 USA

                trevino@lakeforest.edu

Abstract

For a positive integer r, define j_{r} to be the smallest positive integer n satisfying n! > n^{r−1}. In this paper we prove j_{r+1} ∈ {j_{r} + 1, j_{r} + 2}, which leads us to explore the set of positive integers r for which j_{r+1} = j_{r} + 2. We prove this set has the same density as the prime numbers. The sequence j_{r} was introduced by Carlson, Goedhart, and Harris in their work on factoradic happy numbers, and we prove some properties of j_{r} that lead to an improvement of one of their theorems.

1 Introduction

Let r be a positive integer and define j_{r} to be the smallest positive integer n satisfying

n! > n^{r−1}.  (1)

The first 20 values of j_{r} (\underline{A230319} in the On-Line Encyclopedia of Integer Sequences [9]) are

{2, 3, 4, 6, 7, 8, 10, 11, 12, 14, 15, 16, 18, 19, 20, 22, 23, 24, 25, 27}.  (2)

The sequence j_{r} made an appearance in work by Carlson, Goedhart, and Harris [2] on factoradic happy numbers. Our main goal in this paper is to prove some properties of j_{r} to be able to improve some of their results.

To get a better background, we need to define happy numbers (\underline{A007770}) and factoradic expansion. Let n be a positive integer and S_{2}(n) be the sum of the squares of its decimal digits. Consider the sequence of iterates of S_{2} on n, i.e., n, S_{2}(n), S_{2}^{2}(n), . . .. It is well known [7, pp. 74, 83–84] that eventually all the terms in the sequence are 1 or eventually the sequence becomes periodic with the cycle

4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4.

If the sequence reaches 1, we say n is happy. Many generalizations of happy numbers have been studied. For example, one can allow to change the base to b ≥ 2 instead of 10, and one can replace sum of squares of digits, with the sum of r-th powers of the digits for some integer r ≥ 1. Let S_{r,b}(n) be the sum of r-th powers of the digits of n when n is written in base b. In depth analysis of the cases r ∈ {2, 3}, b ∈ {2, 3, …, 10} has been done by Grundman and Teeple [4]. The techniques developed by Grundman and Teeple [4] can easily be used to study other choices of r and b. Another generalization is to allow the base b to be a negative number. This has been done by Grundman and Harris [6] for −2 ≥ b ≥ −10 and r = 2. The authors also study in what cases there exist consecutive b-happy numbers in an arithmetic progression, generalizing work of El-Sedy and Siksek [3] (who do this for happy numbers) and the work of Grundman and Teeple [5] (who do this for b-happy r-power happy numbers). Bland, Cramer, de Castro, Domini, Edgar, Johnson, Klee, Koblitz, and Sundaresan [1] addressed a series of questions regarding a generalization of happy numbers to the fractional base 3/2, and Treviño and Zhylinski [10] addressed other fractional bases.

Every positive integer n can be written uniquely in the form

n = ∑_{i=1}^{k} a_{i} · i!,

for some positive integer k satisfying 1 ≤ a_{k} ≤ k, and 0 ≤ a_{i} ≤ i for 1 ≤ i ≤ k−1. We call this the factoradic expansion of n. We use the notation n = (a_{k}a_{k−1} ⋯ a_{1})_{!} to express a number written in its factoradic expansion. For example, 8 = 110_{!} because 8 = 0 · 1! + 1 · 2! + 1 · 3!.

Carlson, Goedhart, and Harris [2] generalized the concept of happy numbers to factoradic expansions as follows: let S_{r,!}(n) be the sum of the r-th powers of the factoradic digits of a number n, then a positive number n is an r-power factoradic happy number if there exists an integer k such that S_{r,!}^{k}(n) = 1 (the k-iteration of S_{r,!} is 1). Their main theorem is that for r ∈ {1, 2, 3, 4}, there exist arbitrarily long sequences of consecutive r-power factoradic happy numbers.

When studying happy numbers, regardless of the setting, it is important to show that the relevant happy function S (S = S_{r,b} or S = S_{r,!}) satisfies S(n) < n for all n > N, for some integer N. One of the difficulties of the factoradic case is the bound for N is less easy to get that in the other happy number generalizations. Carlson, Goedhart, and Harris [2, Theorem 10] proved that for 2 ≤ r ≤ 30, they can choose N = M_{r} = ∑_{i=1}^{j_{r}} i · i! = (j_{r} + 1)! − 1. Our main motivation to study j_{r} is to be able to prove this result for all r, i.e., to prove

Theorem 1. Let r be a positive integer. Write n in its factoradic expansion as n = ∑_{i=1}^{k} a_{i}i! with 1 ≤ a_{k} ≤ k, and 0 ≤ a_{i} ≤ i for i ∈ {1, 2, …, k − 1}. Let

S_{r,!}(n) = ∑_{i=1}^{k} a_{i}^{r}.

Then for n ≥ (j_{r} + 1)!,

S_{r,!}(n) < n.

In our quest to prove the above theorem, we need to study properties of the sequence j_{r}. In Section 2 we prove some properties of j_{r} that study how the sequence grows. For example, we prove that j_{r+1} − j_{r} ∈ {1, 2}. We also prove that the number of positive integers r ≤ x for which j_{r+1} − j_{r} = 2 is asymptotic to x/ log x. Inspired by this, we say r is a j-prime if j_{r+1} − j_{r} = 2. In Section 3, we prove some lemmas that are necessary for our proof of Theorem 1, in particular we have a nice upper bound for sums of powers. Finally, in Section 4, we prove Theorem 1.

2 Studying the sequence j_{r}

For the purposes of proving Theorem 1, we need an upper bound on j_{r}. The following proposition is the main result from this section we need.

Proposition 2. Let ε > 0 be a real number. Then there exists M such that, for integers r > M, we have that j_{r} < (1 + ε)r.

Proof. It is enough to prove that there exists an M such that, for r > M,

log(⌊(1 + ε)r⌋!) > (r − 1) log⌊(1 + ε)r⌋.

By expanding log as a sum we have

log(⌊(1 + ε)r⌋!) > ∫_{1}^{⌊(1+ε)r⌋} log t dt = ⌊(1 + ε)r⌋ log⌊(1 + ε)r⌋ − ⌊(1 + ε)r⌋ + 1.

Now, we want to show that there exists M such that, for r > M,

⌊(1 + ε)r⌋ log⌊(1 + ε)r⌋ − ⌊(1 + ε)r⌋ + 1 > (r − 1) log⌊(1 + ε)r⌋,

which is equivalent to

log⌊(1 + ε)r⌋ > \frac{⌊(1 + ε)r⌋ − 1}{⌊(1 + ε)r⌋ − r + 1}.

Since ⌊(1 + ε)r⌋ > (1 + ε)r − 1, then

log⌊(1 + ε)r⌋ > log((1 + ε)r − 1),

and

\frac{⌊(1 + ε)r⌋ − 1}{⌊(1 + ε)r⌋ − r + 1} = 1 + \frac{r − 2}{⌊(1 + ε)r⌋ − r + 1} < 1 + \frac{r − 2}{εr}.

If we find an M such that, for r > M, we have

log((1 + ε)r − 1) > 1 + \frac{r − 2}{εr},

then we are done. Note that

lim_{r→∞} log((1 + ε)r − 1) = ∞ and lim_{r→∞} (1 + \frac{r − 2}{εr}) = 1 + \frac{1}{ε}

since ε > 0. This clearly implies the existence of the desired M.

Remark 3. Given ε as in Proposition 2, we can take M to be

M = ⌊\frac{e^{1+ \frac{1}{ε}} + 1}{1 + ε}⌋ .

Then, for r > M, we have that r > \frac{e^{1+ \frac{1}{ε}} +1}{1+ε}. Thus,

log ((1 + ε)r − 1) > 1 + \frac{1}{ε} > 1 + \frac{1}{ε} − \frac{2}{εr} = 1 + \frac{r − 2}{εr},

where the last inequality holds since ε > 0.

The following theorem provides an asymptotic for j_{r} that improves the upper bound from Proposition 2 and provides a strong lower bound. We do not need this result to prove our main theorem, but the result might be of independent interest.

Theorem 4. For a positive integer r, there exists a real number θ_{r} such that

j_{r} = r + \frac{r}{log r} + θ_{r} (\frac{r}{log r}) ,

with θ_{r} → 0 as r → ∞.

Proof. We first prove the upper bound. Let ε > 0 be a real number. We show that

j_{r} < r + \frac{(1 + ε)r}{log r}

for r sufficiently large. It suffices to prove that

∑_{i=1}^{r+⌊\frac{(1+ε)r}{log r}⌋} log i > (r − 1) log (r + ⌊\frac{(1 + ε)r}{log r}⌋)  (3)

for r sufficiently large. Let n = r + ⌊\frac{(1+ε)r}{log r}⌋. We can bound the left side of (3) using the following explicit version of the lower bound in Stirling’s formula [8], which is valid for every positive integer n:

∑_{i=1}^{n} log i > \frac{1}{2} log (2π) + (n + \frac{1}{2}) log n − n + \frac{1}{12n + 1}.  (4)

Applying (4) to (3), it suffices to show

log n > 1 + \frac{r − \frac{3}{2}}{⌊\frac{(1+ε)r}{log r}⌋ + \frac{3}{2}} − \frac{log (2π)}{2 (⌊\frac{(1+ε)r}{log r}⌋ + \frac{3}{2})} − \frac{1}{(12n + 1) (⌊\frac{(1+ε)r}{log r}⌋ + \frac{3}{2})}.

Since \frac{(1+ε)r}{log r} ≥ ⌊\frac{(1+ε)r}{log r}⌋ > \frac{(1+ε)r}{log r} − 1, we need only prove

log (r + \frac{(1 + ε)r}{log r} − 1) > 1 + (\frac{1}{1 + ε}) log r.

But the left side is greater than log r which is larger than 1 + \frac{1}{1+ε} log r for r sufficiently large. For the lower bound, we prove that

j_{r} > r + \frac{r}{log r}

for r sufficiently large. It suffices to prove that

∑_{i=1}^{r+⌊\frac{r}{log r}⌋} log i ≤ (r − 1) log (r + ⌊\frac{r}{log r}⌋)  (5)

for r sufficiently large.

Let n = r + ⌊\frac{r}{log r}⌋. We can bound the left side of (5) using the following explicit version of the upper bound in Stirling’s formula [8], which is valid for every positive integer n:

∑_{i=1}^{n} log i < \frac{1}{2} log (2π) + (n + \frac{1}{2}) log n − n + \frac{1}{12n}.  (6)

Applying (6) to (5), we need only show

log n ≤ 1 + \frac{r − \frac{3}{2}}{⌊\frac{r}{log r}⌋ + \frac{3}{2}} − \frac{log (2π)}{2 (⌊\frac{r}{log r}⌋ + \frac{3}{2})} − \frac{1}{12 (r + ⌊\frac{r}{log r}⌋) (⌊\frac{r}{log r}⌋ + \frac{3}{2})}.

Since \frac{r}{log r} ≥ ⌊\frac{r}{log r}⌋ > \frac{r}{log r} − 1, letting x = r + \frac{r}{log r}, we want to show that

log x ≤ 1 + \frac{r − \frac{3}{2}}{\frac{r}{log r} + \frac{3}{2}} − \frac{log (2π)}{2 (\frac{r}{log r} + \frac{1}{2})} − \frac{1}{12 (x − 1) (\frac{r}{log r} + \frac{1}{2})}

for r sufficiently large; or, after subtracting by log r on both sides,

log (1 + \frac{1}{log r}) ≤ 1 − \frac{3(log^{2} r + log r)}{2r + 3 log r} − \frac{log (2π)}{2 (\frac{r}{log r} + \frac{1}{2})} − \frac{1}{12 (x − 1) (\frac{r}{log r} + \frac{1}{2})}.

Note that the left side goes to 0 as r → ∞ while the right side goes to 1 as r → ∞. Therefore the inequality is true for r sufficiently large.

In the interest of exploring j_{r} further, let’s analyze the relationship between j_{r} and j_{r+1}. Proposition 5. For all positive integers r we have that j_{r} + 1 ≤ j_{r+1} ≤ j_{r} + 2. Proof. It is easy to see that j_{r+1} > j_{r}. So, it suffices to prove that

(j_{r} + 2)! > (j_{r} + 2)^{r}.

Assume, for the sake of contradiction, that (j_{r} + 2)! ≤ (j_{r} + 2)^{r}. Then,

\frac{1}{(j_{r} + 2)!} ≥ \frac{1}{(j_{r} + 2)^{r}} .  (7)

Multiplying inequalities (1) and (7),

\frac{j_{r}!}{(j_{r} + 2)!} > \frac{j_{r}^{r−1}}{(j_{r} + 2)^{r}} .

It follows that,

\frac{1}{j_{r} + 1} > \frac{j_{r}^{r−1}}{(j_{r} + 2)^{r−1}} = (\frac{j_{r}}{j_{r} + 2})^{r−1} ,

which implies

(1 + \frac{2}{j_{r}} )^{r−1} > j_{r} + 1.  (8)

Since r^{r−1} ≥ r(r − 1) ⋯ (2) = r!, we know j_{r} ≥ r. Using this in (8),

(1 + \frac{2}{r})^{r−1} ≥ (1 + \frac{2}{j_{r}} )^{r−1} > j_{r} + 1 ≥ r + 1.

Therefore

(1 + \frac{2}{r})^{r−1} > r + 1.  (9)

But (1 + \frac{2}{r})^{r−1} ≤ e^{2} < 8. This contradicts (9) for r ≥ 7. We can use (2) to confirm that (9) is not satisfied for 1 ≤ r ≤ 6. Thus (9) fails for r ≥ 1. The result follows.

Therefore, for some positive integers r we have j_{r+1} = j_{r} + 1 and for others we have j_{r+1} = j_{r} + 2. Let S be the set of positive integers such that j_{r+1} = j_{r} + 2. That is

S = {r ∈ ℕ | j_{r+1} = j_{r} + 2}.  (10)

The first few values in S can be easily determined from (2) to be 3, 6, 9, 12, 15, 19. This is sequence \underline{A336803}. From Theorem 4 and Proposition 5 we can prove the following theorem which inspired us to name the elements of S as j-primes.

Theorem 6. For a positive real number x, let S(x) = {r ≤ x | j_{r+1} = j_{r} + 2}. Then

S(x) ∼ \frac{x}{log x}.

Proof. From Proposition 5 we know j_{r} ∈ {j_{r−1} + 1, j_{r−1} + 2}. Therefore

j_{r} = j_{1} + r − 1 + ∑_{\begin{matrix}i∈S \\ i≤r \end{matrix}} 1 = r + 1 + S(r).

From Theorem 4 we have j_{r} − r ∼ \frac{r}{log r}. Hence S(r) ∼ \frac{r}{log r}.

3 Some useful lemmas

We are interested in the sum of the first n = j_{r} r-powers. There is an easy way to get a bound that works for all n:

∑_{i=1}^{n} i^{r} ≤ ∫_{1}^{n+1} t^{r} dt = \frac{(n + 1)^{r+1} − 1}{r + 1}.  (11)

For large n, (11) is great, but for n close to r, which is more relevant when n = j_{r}, we need a tighter bound. First, we recall some known properties of the Bernoulli numbers (\underline{A027641}), where we take B_{1} = \frac{1}{2} instead of the usual B_{1} = −\frac{1}{2} to simplify (14).

• For k > 0, B_{k} < 0 if and only if k ≡ 0 (mod 4).

• We have

∑_{k=0}^{∞} \frac{B_{k}}{k!} = \frac{e}{e − 1}.  (12)

• For k > 0, we have

B_{2k} = \frac{(−1)^{k+1}2(2k)!}{(2π)^{2k}}ζ(2k).  (13)

• B_{2k+1} = 0 for k ≥ 1.

• (Bernoulli’s formula) For positive integers n and r, we have

∑_{i=1}^{n} i^{r} = \frac{1}{r + 1} ∑_{k=0}^{r} \binom{r + 1}{k}B_{k}n^{r+1−k}.  (14)

Lemma 7. Let r ≥ 2 be an integer. Then, for an integer n ≥ r + 1,

∑_{i=1}^{n} i^{r} ≤ \frac{Cn^{r+1}}{r + 1}

where C = \frac{e}{e−1} + \frac{π^{4}}{45(16π^{4}−1)} = 1.583 · · ·.

Proof. For n ≥ r + 1, observe that \binom{r+1}{k} ≤ \frac{(r+1)^{k}}{k!} ≤ \frac{n^{k}}{k!} . Using that B_{2k} < 0 for even k ≥ 1 and B_{k} = 0 for odd k > 1, by Bernoulli’s formula (14) we have

∑_{i=1}^{n} i^{r} ≤ \frac{1}{r + 1} (1 + ∑_{\begin{matrix}1≤k≤r \\ k ≢ 0 (mod 4) \end{matrix}} \frac{n^{k}}{k!} B_{k}n^{r+1−k}) ≤ \frac{n^{r+1}}{r + 1} (1 + ∑_{\begin{matrix}1≤k≤r \\ k ≢ 0 (mod 4) \end{matrix}} \frac{B_{k}}{k!} ) .

Now, using (12), (13), and using ζ(4k) ≤ ζ(4) = \frac{π^{4}}{90} , we have

1 + ∑_{\begin{matrix}1≤k≤r \\ k ≢ 0 (mod 4) \end{matrix}} \frac{B_{k}}{k!} ≤ ∑_{k=0}^{∞} \frac{B_{k}}{k!} − ∑_{k=1}^{∞} \frac{B_{4k}}{(4k)!}

= \frac{e}{e − 1} + 2 ∑_{k=1}^{∞} \frac{1}{(2π)^{4k}} ζ(4k)
≤ \frac{e}{e − 1} + 2ζ(4)\frac{1}{(2π)^{4} (1 − \frac{1}{(2π)^{4}} )}

= \frac{e}{e − 1} + \frac{π^{4}}{45(16π^{4} − 1)}.

Therefore, for n ≥ r + 1,

∑_{i=1}^{n} i^{r} ≤ C \frac{n^{r+1}}{r + 1},

where C = \frac{e}{e−1} + \frac{π^{4}}{45(16π^{4}−1)}, as desired.

What follows is a simple inequality we need to use to prove our main theorem.

Lemma 8. Let 0 < x < r be a real number. Then, for all integers r ≥ 1, we have that

(1 − \frac{x}{r})^{r} < e^{−x}.

Proof. Using the Taylor series expansion of log(1 − x/r), we have that

r log (1 − \frac{x}{r}) = −r (\frac{x}{r} + \frac{1}{2} (\frac{x}{r})^{2} + \frac{1}{3} (\frac{x}{r})^{3} + ⋯) = −x − \frac{x^{2}}{2r} − \frac{x^{3}}{3r^{2}} − ⋯ < −x.

Lemma 9. Let 0 < ε < 2. Then there exists a positive integer M such that, for all integers r > M, we have that

(1 − \frac{1}{j_{r}} )^{r+1} < e^{−1+ε}.

Proof. Let ε_{1} > 0 be a real number. First, by Proposition 2, there exists M_{1} positive integer such that, for r > M_{1}, we have that j_{r} < (1 + ε_{1})r. Then, for r > M_{1}, we have

(1 − \frac{1}{j_{r}} )^{r+1} < (1 − \frac{1}{(1 + ε_{1})r})^{r+1} .  (15)

Observe that

lim_{r→∞} (1 − \frac{1}{(1 + ε_{1})r})^{r+1} = e^{− \frac{1}{1+ε_{1}}} .  (16)

On the other hand, we know that j_{r} > r because j_{1} > 1 and j_{r+1} ≥ j_{r} + 1. It follows that

(1 − \frac{1}{j_{r}} )^{r+1} > (1 − \frac{1}{r})^{r+1} ,  (17)

and we have that

lim_{r→∞} (1 − \frac{1}{r})^{r+1} = e^{−1}.  (18)

Since (15) and (16) hold for all ε_{1} > 0, from (15), (16), (17), and (18) we conclude that

lim_{r→∞} (1 − \frac{1}{j_{r}} )^{r+1} = e^{−1}.

Thus, there exists M such that, for r > M, we have that

(1 − \frac{1}{j_{r}} )^{r+1} < e^{−1+ε}

as desired. Remark 10. From Remark 3, we know that

j_{r} < (\frac{1}{1 − \frac{ε}{2}} ) r

for r > (e^{\frac{2}{ε}} + 1) (1 − \frac{ε}{2}). Then,

(1 − \frac{1}{j_{r}} )^{r+1} < (1 − \frac{1 − \frac{ε}{2}}{r} )^{r+1} < (1 − \frac{1 − \frac{ε}{2}}{r} )^{r}

where

(1 − \frac{1 − \frac{ε}{2}}{r} )^{r} < e^{−1+ε}

for all integers r ≥ 1 (by Lemma 8 with x = 1 − ε). Therefore, we can take M in Lemma 9 as

M = ⌊(e^{\frac{2}{ε}} + 1) (1 − \frac{ε}{2})⌋ .

4 Proof of Theorem 1

Carlson, Goedhart, and Harris [2, Theorem 10] reduced the proof of Theorem 1 to showing the inequality^{1}

(j_{r} + 1)! − (j_{r} + 1)^{r−1} + j_{r}! − 1 − ∑_{i=1}^{j_{r}−1} i^{r} > 0.  (19)

They proved it via computation [2, Table 1] for 2 ≤ r ≤ 30. We prove that (19) works for all positive integers r in the next proposition.

Proposition 11. Let r be a positive integer. Then

(j_{r} + 1)! − (j_{r} + 1)^{r−1} + j_{r}! − 1 − ∑_{i=1}^{j_{r}−1} i^{r} > 0.

Proof. Let C = \frac{e}{e−1} + \frac{π^{4}}{45(16π^{4}−1)} and let L_{r} be the left side of (19). We want to show L_{r} > 0. We know that j_{r} > r. Also, as observed by Carlson, Goedhart, and Harris [2], we have that

(j_{r} + 1)^{r−1} = j_{r}^{r−1} (1 + \frac{1}{j_{r}} )^{r−1} < j_{r}^{r−1} (1 + \frac{1}{r})^{r} < 3j_{r}^{r−1}.

From this and from Lemma 7, we have that for r ≥ 2,

L_{r} > j_{r}!(j_{r} + 2) − 3j_{r}^{r−1} − 1 − \frac{C(j_{r} − 1)^{r+1}}{r + 1}

= j_{r}!(j_{r} + 2) − 3j_{r}^{r−1} − 1 − j_{r}^{r−1} (\frac{j_{r}^{2}}{r + 1}) (C (\frac{j_{r} − 1}{j_{r}})^{r+1})

> j_{r}^{r−1} (r − 1 − (\frac{j_{r}^{2}}{r + 1}) (C (1 − \frac{1}{j_{r}} )^{r+1})) − 1.

Let 0 < ε < 1 − log(C) be a real number. By Lemma 9, there exists a positive integer M_{1} such that, for r > M_{1},

(1 − \frac{1}{j_{r}} )^{r+1} < e^{−1+ε},

which leads to

C (1 − \frac{1}{j_{r}} )^{r+1} < Ce^{−1+ε}.

1They incorrectly omitted the “−1” term from the inequality.

It follows that, for r > max(2, M_{1}),

L_{r} > j_{r}^{r−1} (r − 1 − Ce^{−1+ε} (\frac{j_{r}^{2}}{r + 1})) − 1

= Ce^{−1+ε}j_{r}^{r−1} ((\frac{1}{Ce^{−1+ε}} ) (r − 1) − \frac{j_{r}^{2}}{r + 1}) − 1.

Let k > 1 be a constant such that k^{2} < \frac{1}{Ce^{−1+ε}}. By Proposition 2, there exists a positive integer M_{2} such that, for r > M_{2}, we have that

j_{r} < k(r + 1).

Then, for r > max(2, M_{1}, M_{2}),

L_{r} > Ce^{−1+ε}j_{r}^{r−1} ((\frac{1}{Ce^{−1+ε}} − k^{2}) r − \frac{1}{Ce^{−1+ε}} − k^{2}) − 1

which is greater than −1 whenever r > \frac{1+Ck^{2}e^{−1+ε}}{1−Ck^{2}e^{−1+ε}}. Since the expression on the left is an

integer, we conclude the result for r > max (2, M_{1}, M_{2}, ⌊\frac{1+Ck^{2}e^{−1+ε}}{1−Ck^{2}e^{−1+ε}} ⌋).

From Remarks 3 and 10 we know that we can choose M_{1}, M_{2} to be

M_{1} = ⌊(e^{\frac{2}{ε}} + 1) (1 − \frac{ε}{2})⌋ ,
M_{2} = ⌊\frac{e^{1+ \frac{1}{k−1}} + 1}{k} − 1⌋ .

The idea is to minimize the value of max (2, M_{1}, M_{2}, ⌊\frac{1+Ck^{2}e^{−1+ε}}{1−Ck^{2}e^{−1+ε}} ⌋) . Choosing ε = 0.26 and k = 1.15, we have that

max (2, M_{1}, M_{2}, ⌊\frac{1 + Ck^{2}e^{−1+ε}}{1 − Ck^{2}e^{−1+ε}} ⌋) = max(2, 1907, 1857, 2167) = 2167.

This proves the result for r > 2167. The cases r ≤ 2167 can be easily checked with a computer.

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[10] E. Treviño and M. Zhylinski, On generalizing happy numbers to fractional-base number systems, Involve 12 (2019), 1143–1151.

2010 Mathematics Subject Classification: Primary 11B83; Secondary 11N37, 11A67, 11B68. Keywords: happy factoradic number, asymptotic growth, sum of powers.

(Concerned with sequences \underline{A007770}, \underline{A027641}, \underline{A230319}, and \underline{A336803}.)

Received January 26 2021; revised versions received January 27 2021; July 20 2021. Published in Journal of Integer Sequences, August 17 2021.

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