5

This comes up in OEIS as A007018. However the recursive form is useless to me, I need the closed form. I've been trying for several hours and I simply come up empty. Any advice?

Thanks.

TechnoSam
  • 375
  • What exactly do you mean with the explicit form? The recursive form looks pretty explicit. Do you mean the closed form to evaluate $a_n$ for any $n$? – Newb Apr 01 '15 at 04:58
  • 2
    The fact that the only closed form given in the OEIS entry is $a(n) = \left\lfloor c^{2^n}\right\rfloor$, where $c=1.597910218031873178338070118157\ldots$, suggests that you won’t do any better than this. – Brian M. Scott Apr 01 '15 at 04:59
  • 1
    @BrianM.Scott Note that $c$ was probably computed by evaluating the sequence recursively up to some $n$. (Similar to the "closed form" for the Fibonacci sequence that requires finding the Golden Ratio to arbitrary precision...) – Newb Apr 01 '15 at 05:00
  • @Newb: Yes, after showing that there is a closed expression of that form. – Brian M. Scott Apr 01 '15 at 05:01
  • @Newb, Yes, my bad, thank you! I'm looking for something like Brian M. Scott posted, but I don't know how I would ever arrive at that solution on my own without a computer. – TechnoSam Apr 01 '15 at 12:08
  • Knuth's comment in OEIS that "Using the methods of Aho and Sloane, Fibonacci Quarterly 11 (1973), 429-437, it is easy to show that $a_n$ is the integer just a tiny bit below the real number $\theta^{2^n}-\frac12$, where $\theta \approx 1.597910218$ is the exponential of the rapidly convergent series $\ln\frac32+\sum_{n \ge 0} \ln(1+(2a_n+1)^{-2})$" gives you a starting point. – Peter Taylor Apr 02 '15 at 08:10

1 Answers1

2

There are just two (diagonalized monic) versions that give nice closed forms: $$ x_n = x_{n-1}^2 $$ gives $$ x_n = x_0^{\left( 2^n \right) }. $$ The other, used by Lucas, is $$ x_n = x_{n-1}^2 - 2; \; \; \; \; x_0 > 2. $$ This time we find $A > 1,$ with $AB = 1$ and $A + B = x_0.$ Then $$ x_n = A^{\left( 2^n \right) } + B^{\left( 2^n \right) }. $$

That is all the nice ones. For the others, taking logarithm of both sides shows that there is a limit which is the number $c$ from OEIS, but we can only estimate $c$ by calculating many terms of the sequence itself.

With your sequence, taking $$ a_n = b_n - \frac{1}{2} $$ gives $$ b_n = b_{n-1}^2 + \frac{1}{4} $$ so you are out of luck as far as closed form answers. With this much, not difficult to prove that $$ \frac{\log b_n}{2^n} $$ has a limit as $n \rightarrow \infty,$ call it $w,$ then the number $c = e^w.$

Will Jagy
  • 146,052