R- continued fraction

We can define a recursive function f to calculate the contiuned fraction

f <- function(v) {
  ifelse(length(v) == 1, v, v[1] + 1 / f(v[-1]))
}

However, a drawback is that there is a limitation on the recursion depth, e.g.,

> f(1:1e5)
Error: node stack overflow

Thus, if you have a large array v, a better option might be using for loops, e.g.,

f <- function(v) {
  if (length(v) == 1) {
    return(v)
  }
  s <- tail(v, 1)
  for (k in (length(v) - 1):1) {
    s <- v[k] + 1 / s
  }
  s
}

and you will see

> f(1:1e5)
[1] 1.433127

Using recursion one option to achieve your desired result may look like so:

cfrac <- function(x) {
  n <- length(x)
  if (n > 1) {
    res <- x[1] + 1 / cfrac(x[2:n]) 
  } else {
    res <- x
  }
  return(res)
}

identical(cfrac(2), 2)
#> [1] TRUE
identical(cfrac(c(2, 3)), 2 + 1 / 3)
#> [1] TRUE
identical(cfrac(c(-3, 2, 18)), -3 + 1 / (2 + 1 / 18))
#> [1] TRUE

Tools going in both directions are in the package contfrac