0

The most famous (and simplest non-trivial) recurrence is the Fibonacci recurrence $F(n)=F(n-1)+F(n-2)$ with $F(0)=0, F(1)=1$. What if we consider instead division based recurrences, the simplest non-trivial one being say: $F(x)=F(x/2)+F(x/3)$ (where now the initial conditions are $F(x)=1$ for $x\in [1,2)$; alternatively/equivalently, the integer based recurrence $F(n) = F(\lfloor \frac n2\rfloor) + F(\lfloor \frac n3\rfloor)$)?

Instead of the "short range" dependencies of the subtraction based, the division based one is more "long range"; i.e. in the subtraction based version, one only needs to keep track of 3 "variables" [in the computer/code sense] $F(k-2), F(k-1), F(k)$ and update them accordingly to get to arbitrary $F(n)$), but the division based one basically needs to keep track of the whole array $F(1),\ldots, F(k)$. The "short range" nature of the subtraction based version makes it possible to use linear algebra techniques to analyze the problem, and lead to closed form formulas for those recurrences.

Has this type of problem been studied before? What progress can be made? (e.g. closed form formulas, asymptotics, bounds, etc.)

D.R.
  • 10,556
  • 1
    Why do you think your two equations $F(x) = F(x/2) + F(x/3)$ and $F(n) = F(\lfloor n/2 \rfloor) + F(\lfloor n/3 \rfloor)$ are equivalent? They seem very different to me. – Robert Israel Jun 27 '24 at 01:22
  • perhaps the function $F(x) = \ln (G(x) )$ is interesting – hellofriends Jun 27 '24 at 01:23
  • For $F(x) = F(x/2) + F(x/3)$ try the equivalent problem $G(y) = G(y-\ln 2)+G(y-\ln 3)$. Then perhaps it is relevant that $\ln 2, \ln 3$ are incommensurable. – GEdgar Jun 27 '24 at 01:26
  • 1
    For $F(n) = F(\lfloor n/2 \rfloor) + F(\lfloor n/3 \rfloor)$ with $F(0) = 1$, see OEIS sequence A088468. – Robert Israel Jun 27 '24 at 01:26
  • @RobertIsrael I think they are equivalent with the resp. initial conditions of $F(x)=1_{[1,2)}(x)$ for $x<2$ and the same thing but restricted to $x=0,1$? – D.R. Jun 27 '24 at 03:19
  • @RobertIsrael also, thanks for the OEIS link. It had a pointer to an Erdos problem that has some beautiful answers on MSE https://math.stackexchange.com/questions/487957/how-prove-this-nice-limit-lim-limits-n-to-infty-fraca-nn-frac12-lo – D.R. Jun 27 '24 at 04:54
  • I'm not familiar with this, but for an asymptotic analysis: https://en.wikipedia.org/wiki/Master_theorem_(analysis_of_algorithms) – Qiaochu Yuan Jun 27 '24 at 06:20

1 Answers1

1

Consider $F(x) = F(x/2) + F(x/3)$. With just $F(x) = 1$ on $[1,2)$, you don't determine $F(x)$ on $[2,3)$ because you need $F(x/3)$. But if you say $F(x) = 0$ for $x < 1$, that makes $F(x) = 1$ on $[2,3)$. Now $F$ will be constant on intervals $[x_j, x_{j+1})$ where $x_j$ are the $3$-smooth positive integers (OEIS sequence A003586, i.e. those whose prime factors are all $\le 3$. If $x_j$ is a $3$-smooth positive integer, $\lfloor x/2 \rfloor = x/2$ and $\lfloor x/3 \rfloor = x/3$, and both of those are also $3$-smooth. So the values taken at these points are the same as for the recursion $F(x) = F(\lfloor x/2 \rfloor) + F(\lfloor x/3 \rfloor)$ with $F(0) = 0$ and $F(1) = 1$.

Robert Israel
  • 470,583