Let $x(1) = 1, x(k+1) = ax(k)+ bk, y(k) = -1 +a^{-k}x(k)$. This recursion seems quite intractable, but it has some interesting features and it is much more friendly than it looks at first glance. Here $a, b$ are two positive parameters, with $a>1$. Let $g(a,b) =\lim_{k\rightarrow\infty} y(k)$.

It is probably easy to establish that $g(a, b) = b f(a)$ where $f(a)$ is a smooth function. Values of $f(a)$ for various $a$'s are pictured below. It is possible that if $a$ is a rational number, then $f(a)$ is also a rational number? This happens frequently. More specifically, my questions are:


  • Find a good approximation for $f(a)$.
  • Or can you find an exact formula for $f(a)$?
  • Is $f(8)$ a rational number?

The last question is very, very important to me as it has potential applications in finding a standard mathematical constant which is a normal number. My wish is that $f(8)$ is not a rational number, though it seems at first glance that the contrary should be true, unfortunately.

enter image description here

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    $\begingroup$ Mathematica yields $x(k)= \dfrac{(b-2) a^{k+1}+a^{k+2}+a^k-a^2 b k+a b (k-1)}{(a-1)^2 a}.$ The relevant command is RSolve[{x[k+1]==a x[k]+b k,x[1]==1},x[k],k]. $\endgroup$ – Adrian Keister Aug 6 at 20:43
  • $\begingroup$ @AdrianKeister that can be slightly simplified by dividing by $a$ top and bottom to get $x(k)= \dfrac{(b-2) a^{k}+a^{k+1}+a^{k-1}-a b k+b (k-1)}{(a-1)^2}$. $\endgroup$ – automaticallyGenerated Aug 6 at 20:47
  • $\begingroup$ Let me re-check my computations. I'm sure yours are correct, but it does not match my results, most likely my computations are not correct. I'll check with $b=1$. $\endgroup$ – Vincent Granville Aug 6 at 21:15
  • $\begingroup$ My formula is wrong. It should be $x(k+1) = ax(k) + b(k+1)$ instead. Let me re-do my computations to match what I posted here, that is, $x(k+1) = ax(k) + bk$. Using one or the other is not important in the end ($f(8) will be rational anyway) and both probably have an exact form. $\endgroup$ – Vincent Granville Aug 6 at 21:35

As stated in the comments there is a closed form for $x(k)$ and hence $y(k)$, $g(a,b)$ etc. We have the solutions $$x(k)=\frac{(b-2)a^k+a^{k+1}+a^{k-1}-abk+b(k-1)}{(a-1)^2}$$ $$y(k)=-1+\frac{b-2+a+a^{-1}-a^{1-k}bk+a^{-k}b(k-1)}{(a-1)^2}$$ Hence as $k\to\infty$ we have that $$g(a,b)=-1+\frac{b-2+a+a^{-1}}{(a-1)^2}=\frac{b}{(a - 1)^2}+\frac1a-1$$ $$f(a,b)=\frac1{(a - 1)^2}+\frac1{ab}-\frac1b$$ So clearly when $a,b\in\mathbb{Q}$ we have that $g(a,b)$ and $f(a,b)$ are rational.

  • $\begingroup$ There were some errors in my own computations, but your answer provide the answer I wanted to have, thus I accepted it. It really solves the deeper question I had in mind anyway, and just as I feared, in a way that that makes my wish not come true. $\endgroup$ – Vincent Granville Aug 6 at 22:21

Mathematica yields $x(k)= \dfrac{(b-2) a^{k+1}+a^{k+2}+a^k-a^2 b k+a b (k-1)}{(a-1)^2 a}.$ The relevant command is RSolve[{x[k+1]==a x[k]+b k,x[1]==1},x[k],k]. It follows that $$y(k)=\frac{a^{-k-1} \left((b-2) a^{k+1}+a^{k+2}+a^k-a^2 b k+a b (k-1)\right)}{(a-1)^2}-1. $$ From here, you can calculate $$g(a,b)=\lim_{k\to\infty}y(k)=\text{ConditionalExpression}\left[\frac{a^2+a (b-2)+1}{(a-1)^2 a}-1,\left(\frac{1}{(a-1)^2}|b|a\right)\in \mathbb{R}\land \ln (a)>0\right]. $$ Then you have $$f(a)=\frac{g(a,b)}{b}=\text{ConditionalExpression}\left[\frac{\frac{a^2+a (b-2)+1}{(a-1)^2 a}-1}{b},\left(\frac{1}{(a-1)^2}|b|a\right)\in \mathbb{R}\land \log (a)>0\right]. $$ From here, you can see that if the conditions are true, then if $a$ and $b$ are rational, $f(a)$ will certainly be rational, unfortunately.

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    $\begingroup$ If $a=8$, this has $(a-1)^2=49$ in the denominator, and that would explain why the decimal representation doesn't repeat in the first $14$ or so digits, so it's not as easy to tell from its decimal expansion that it's rational $\endgroup$ – J. W. Tanner Aug 6 at 20:54

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