# An intriguing recursion

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:

Questions

• 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.

• 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]. – Adrian Keister Aug 6 '19 at 20:43
• @AdrianKeister that can be slightly simplified by dividing by $a$ top and bottom to get $x(k)= \dfrac{(b-2) a^{k}+a^{k+1}+a^{k-1}-a b k+b (k-1)}{(a-1)^2}$. – Varun Vejalla Aug 6 '19 at 20:47
• 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 $b=1$. – Vincent Granville Aug 6 '19 at 21:15