My goal is to approximate the sequence $A305706$, where it is defined formally as:
$a(k)$ = smallest $n$ such that the sum of digits of $k^n$ is greater than $k$, or $0$ if no such $n$ exists. ($k \geqslant 1.$)
- My approach
[Remark: I am in grade $11$, so I may have misused a lot of mathematical notation, requesting you to be patient with me :) ]
By a theorem of Senge (see this note), only for $k = 10^m$ ($m \in \mathbb{N_0}$), $a(k)$ is undefined. So, for now, let's take $k \neq 10^m$. The number of digits of a positive integer $n \geqslant1$ written in base $10$ is $\lfloor \log_{10} n\rfloor +1$. So, the number of digits of $k^n$ is $\lfloor n \log_{10}k \rfloor+1.$ Let $k = q \cdot 10^{\nu_{10}(k)}$ ($q \in \mathbb{N}$) where $\nu_{10}(n) := \min \{\nu_2(n),\nu_5(n)\}$ and $\nu_p(n)$ refers to p-adic valuation of $n$. Let $D(n) $ be the sum of the decimal digits of $n$ in base $10$. Then, $D(k^n)=D(q^n)$ as multiplying by powers of $10$ only adds trailing zeros. The number of digits that $q^n$ has is $\lfloor n \log_{10}q \rfloor+1 \sim n \log_{10}q.$ We want $n$ such that $D(k^n)>k$, so for an heuristic approximation of $n$, let us consider $D(k^n) \gtrsim k.$ Assuming a uniform distribution of digits in the decimal expansion of $q^n$, one can assume that each digit contributes $$\frac{\sum_{0 \leqslant i \leqslant 9}i}{10}=9/2.$$ in $D(q^n)$. This assumption implies a rough estimate that $$D(q^n) \stackrel{?}{\sim} 4.5 \cdot n \log_{10}q.$$ We wanted $$D(k^n) \gtrsim k \implies 4.5 \cdot n \log_{10}q \stackrel{?}{\gtrsim} k \implies n \stackrel{?}{\gtrsim} \frac{2k}{9\log_{10}q}.$$ Subsitute the definition of $q$, giving $$n \stackrel{?}{\gtrsim} \frac{2k}{9 \log_{10}(k/10^{\nu_{10}(k)})}$$ Therfore, we have $$a(k) \stackrel{?}{\gtrsim} \frac{2k}{9 \log_{10}k-9\min \{\nu_2(k),\nu_5(k)\}}.$$ Obviously, the major issue here is my assumption of a uniform distribution of digits in $q^n$; which prevents me from getting any rigorous data about $a(k)$ from this approach. While I am aware of a procedure to obtain a rigorous lower bound on $a(k)$, I am afraid that that particular lower bound is exceptionally loose, meaning any heuristics and approximations from that approach are pretty useless.
My question. I am wondering whether there is some stronger yet rigorous way of approximating $a(k)$? I think fields like Probabilistic Number Theory could have some approaches..
Thank you.
- Extra information
The following is a graph of $a(k)$ divided by my heuristic approximation.
- Miscellaneous analysis: Rigorous Lower bounds for $a(k)$.
The number of digits that $q^n$ has in base $10$ is $\lfloor n \log_{10} q \rfloor + 1.$ We wanted $D(k^n)=D(q^n)>k.$ Since in base $10$ each digit can have the maximum face value of $9$, we obtain:$$9\lfloor n \log_{10} q \rfloor + 9 \geqslant D(q^n)>k.$$ Therefore, any possible value of $n$ must satisfy $9\lfloor n \log_{10} q \rfloor + 9 >k \implies \lfloor n \log_{10} q \rfloor>\frac{k-9}{9}.$ As $$n\log_{10}q - \{n \log_{10}q\}= \lfloor n \log_{10} q \rfloor \implies n\log_{10}q - \{n \log_{10}q\}>\frac{k-9}{9}. \implies n \log_{10}q>\frac{k-9}{9}+\{n \log_{10}q\}.$$ This gives, $n>\frac{k-9}{9 \log_{10}q}+\frac{\{n \log_{10}q\}}{\log_{10}q}$ as $q \geqslant 2.$ This finally gives us $$a(k)>\frac{k-9}{9 \log_{10}k-9\min \{\nu_2(k),\nu_5(k)\}}.$$
- Miscellanous Analysis - Explicit upper bounds for $a(k)$?
I thank @GerryMyerson for suggesting this paper by C.L. Stewart. By the Theorem of Senge, we note that $a(k)<\infty.$ The problem with getting an explicit upper bound for $a(k)$ is that it requires an explicit lower bound for $D(q^n),$ which is a highly non-trivial problem. This paper by Stewart provides some theoretical machinery that, in theory, can be used to construct explicit upper bounds for $a(k)$.
Let $a,b,t$ be integers greater than one and let $\alpha , \beta \in \mathbb{Z}$ such that $0 \leqslant \alpha < a $ and $ 0 \leqslant \beta < b.$ Denote the number of digits in the canonical expansion of $t$ in base $a$ which are different from $\alpha$ by $L_{\alpha,a}(t).$ Define $L_{\beta,b}(t)$ similarly and define $L_{\alpha,a,\beta,b}(t):=L_{\alpha,a}(t)+L_{\beta,b}(t).$ For our case, we let $a=q$ (note that $q \geqslant 2$ as $q \neq 1$) and $\alpha = 0$. Let $t = q^n$ and consider $L_{0,q}(q^n)$. Note that writing $q^n$ in base $q$ is just $1$ followed by $n$ zeros.
Therefore, $L_{0,q}(q^n)=1.$ Now, let $b=10$ and $\beta = 0$. By definition, $L_{0,10}(q^n)$ is the number of digits of $q^n$ in base $10$ which are different from $0$.
Therefore, $D(q^n) \geqslant L_{0,10}(q^n).$ We are now in a position to apply Stewart's theorem.
Theorem $1$ (C.L. Stewart) in the document states that if $\log a / \log b \notin \mathbb{Q}$, then $$L_{\alpha, a, \beta, b}(t)> \frac{\log \log t}{\log \log \log t + C}-1,$$
for $t>25$ and $C>0$ is a constant effectively computable in terms of $a$ and $b$ only.
We note that $\log q/\log 10 = \log_{10}q \notin \mathbb{Q}$. (As $q \neq 10^j$ for $j \in \mathbb{N}_0$ by the definition of $k$.) Therefore, we have: for $q^n>25$, $$L_{0,q,0,10}(q^n)> \frac{\log \log q^n}{\log \log \log q^n + C}-1.$$
This implies that, $$L_{0,10}(q^n)>\frac{\log \log q^n}{\log \log \log q^n + C}-2.$$
Consider sufficiently large $n$ which satisfy $$D(q^n) \geqslant L_{0,10}(q^n)>\frac{\log \log q^n}{\log \log \log q^n + C}-2>k,$$ and $q^n > 25$. This implies that sufficiently large $n$ satisfying $$\frac{\log \log q^n}{\log \log \log q^n + C}-2>k$$ and $q^n>25$ also satisfy $D(k^n)=D(q^n)>k.$
In conclusion, for a fixed $k$ we have a fixed $q \geqslant 2$, and letting $n=N$ where $N$ is any sufficiently large natural number satisfying $$\frac{\log \log q^N}{\log \log \log q^N + C}-2>k$$ and $q^N>25$ also satisfies $D(k^N)>k \implies N \geqslant a(k).$
Hence, we have obtained an explicit upper bound on $a(k)$.
Further information. Notice that: $$\lim_{N \to \infty}q^N = + \infty$$ and $$\lim_{N \to \infty}\frac{\log \log q^N}{\log \log \log q^N + C}-2 = + \infty.$$ This guarantees that we are always able to find $N$ that satisfies the required conditions.
Remark: This upper bound is obviously extremely weak. For example, it gives that $1.2045589414\times10^{14} > a(7)$. ($a(7) = 2$.) :)
Update 1: Stewart's theorem is probably (?) more effective by using $a = k$ (note that $k \geqslant 2$ as $k \neq 1$.) and $t=k^n$. Of course, $\log_{10}k \notin \mathbb{Q}$ as $k \neq 10^m$ for $m \in \mathbb{N}_0$. Using $k$ instead of $q$ is more effective as $k \geqslant q. \implies k^N \geqslant q^N$.
- Miscellaneous analysis: Applying Radcliffe's result.
In this paper, Radcliffe proved; $D(2^n)> \log_4n $ for all $n \in \mathbb{N}.$ We can apply his result by considering special values of $k$. Particularly, let $k = 2 \cdot 10^\varphi$; $\varphi \in \mathbb{N}_0$.
Therefore, for such $k$, we have $q=2$. Hence, for these values of $k$, $D(k^n)=D(q^n)=D(2^n).$ Consider $N$ such that $\log_{4}N>k \implies N >4^k.$ By Radcliffe's result, $N \geqslant 4^k+1$ satisfy $$D(k^N)=D(2^N)> \log_{4}N>k \implies D(k^N)>k \implies N \geqslant a(k) \implies 4^k+1 \geqslant a(k)$$ for such $k$ having $q=2$.
- Miscellaneous analysis: Using Stewart's Theorem $2$ for explicit upper bounds?
Let $u_n$ be the $n$-th term of a general linear recurrence sequence satisfying $$u_n = d_1u_{n-1}+ \ldots + d_ju_{n-j},$$ where $d_1, \ldots ,d_j$ and $u_1, \ldots , u_j$ are integers. Then, $$u_n=P_1(n)\lambda_1^n+ \ldots +P_k(n)\lambda_k^n,$$ where $\lambda_1,\ldots , \lambda_k$ are the roots of the characteristic polynomial associated with $u_n$.
Theorem $2$ (C.L. Stewart) Let $a$ be an integer larger than one and let $\alpha$ be a non-negative integer less than $a$. If $u_n$ satisfies $=P_1(n)\lambda_1^n+ \ldots +P_k(n)\lambda_k^n$, $|\lambda_1|>\max\{1,|\lambda_2|, \ldots, |\lambda_k|\}, P_1(x)$ is not identically zero, and $\frac{\log \lambda_1}{\log a}$ is irrational then,
$$L_{\alpha, a}(u_n)> \frac{\log n}{\log \log n + C_0}-1,$$ for $n>4$, where $C_0>0$ which is effectively computable in terms of $a$,$\lambda_1, \ldots , \lambda_k$,$j$ and the coefficients of $P_1(x), \ldots , P_k(x)$.
Consider a homogenous linear recurrence sequence satisfying $$x_n = kx_{n-1}+x_{n-2}-kx_{n-3}$$ and let $x_0=3, x_1=k$ and $x_2 = k^2+2$. By a standard procedure, it can be proved that the characteristic polynomial associated with $\{x_n\}_{n \geqslant 0}$ is $(r-1)(r+1)(r-k)=0.$ Let $\lambda_1=k, \lambda_2=1$ and $\lambda_3 = -1.$ By a standard technique, it can be shown that $x_n=1 \cdot \lambda_1^n+1\cdot \lambda_2^n + 1 \cdot \lambda_3^n,$ which implies that $P_1(x)=P_2(x)=P_3(x)=1.$
This implies that $P_1(x)$ is not identically zero. Let $a=10$ and $\alpha =0.$ Obviously, $| \lambda_1| > \max\{1,|\lambda_2|,|\lambda_3|\}$ as $k \geqslant2.$ We know $\frac{\log \lambda_1}{\log a} = \log_{10}k \notin \mathbb{Q}.$ Hence, the hypotheses of Theorem $2$ are satisfied.
Let $N>4$ be any sufficiently large odd natural number satisfying $$\frac{\log N}{\log \log N + C_0}-1>k,$$ then $N \geqslant a(k),$ as by Theorem $2$,
$$D(x_N) \geqslant L_{0,10}(x_N)> \frac{\log N}{\log \log N + C_0}-1$$
and as $N$ is odd, this implies that $x_N=k^N$, therefore giving,
$$D(k^N) \geqslant L_{0,10}(k^N)> \frac{ \log N}{\log \log N + C_0}-1>k,$$
which implies that $D(k^N) > k \implies N \geqslant a(k).$ QED.
Remark: (For the interested reader) A complementary and slightly more general result of Theorem $2$ from Stewart's paper is presented in this paper by Florian Luca.
Update $2$: Going through the presentation slides of a talk by Stefan Porubsky given on June $16$, $2023$, titled "Properties of Sum-of-digits functions", one notes that the content explicitly concerning lower bounds for $D(n)$ (which can be used to find upper bounds for $a(k)$ ) is only within $2$ slides. (Page number $69-70$ or slide number $52$ ) Out of these two slides, one finds the mention of three papers. Particularly, these papers are On the sum of digits of some sequences of integers,On the sum of digits of the factorial, and The Number of Non-Zero Digits of $n!$. Out of these three papers, the first paper yields the most general results; however, I think that most likely (?) the sequence of $a_n = k^n$ does not meet the criteria which is required to invoke Theorem $1.1$ in the first paper..
As suggested by @Gerry, I've also skimmed through some papers by Mauduit and Drmota; however, it seems that most of the papers authored by these authors on this topic concern congruence relations and asymptotic properties of expressions like $\sum D(n)$, which I don't think could be applied to this problem. (?)
- Miscellaneous analysis: Proving new lower bounds for $D(5^n).$
In this rough note, I extend Radcliffe's result of $D(2^n) > \log_4(n)$ to a new result concerning powers of five. Particularly, I prove that $$D(5^n)> \log_{2}(n).$$ It is now obvious that this result immediately implies explicit upper bounds for $a(k)$ when $k$ satisfies $k = 5 \cdot 10^{\varphi}$ for $\varphi \in \mathbb{N}_0.$ In other words, such $k$ where $q = 5$.