Behavour of the rpois function

I'm trying to understand a proposed solution for a University test.
Let me assume that we have created a random variable with

set.seed(123)
R <- 5
X <- rexp(R, 2)

So the content of X is

0.42172863 0.28830514 0.66452743 0.01578868 0.02810549

In the solutions of the problem I find

Y <- rpois(R, exp(X / 4))

where the content of exp(X / 4) is

1.111191 1.074737 1.180729 1.003955 1.007051

where, contrary to my expectations the second argument is an array instead of being a scalar.

If I calculate

print(rpois(R, 1.111191))
print(rpois(R, 1.074737))
print(rpois(R, 1.180729))
print(rpois(R, 1.003955))
print(rpois(R, 1.007051))

I get

2 1 1 3 1
1 1 0 2 0
0 1 3 3 2
1 4 1 1 1
1 0 0 3 2

while for rpois(R, exp(X / 4)) I get

1 2 0 1 2

How are the two results related?

It's a behaviour I can't find explained anywhere.

R makes its functions vectorized wherever it's reasonable to do so.

In particular, in the function call rpois(R, lambda), R specifies the number of samples to take, and lambda is the vector of means, which is recycled to match R. In other words, if lambda is a single value then the same mean will be used for each Poisson draw; if it is a vector of length R, then each element of the vector will be used for the corresponding Poisson draw.

So the equivalent of Y <- rpois(R, exp(X / 4)) would be

Y <- c(
   rpois(1, exp(X[1]/4),
   rpois(1, exp(X[2]/4),
   rpois(1, exp(X[3]/4),
   ...
)

We could also do this with a for loop:

Y <- numeric(R)  ## allocate a length-R numeric vector
for (i in seq(R)) {
  Y[i] <- rpois(1, exp(X[i]/4))
}

Using the vectorized version whenever it's available is best practice; it's faster and requires less code (therefore easier to understand).