Define F(n) as the minimum number of tokens in Pure Factor (A version of Factor where the only syntax is [, and only uses words that are in kernel and deal with only quotations, such as dip and keep, and call has its restrictions removed) required to put n elements on the stack.
Let the number in this post be the minimum n where F(n) = 10^100.
F(n) <= floor(n/4) + 3 (except for n = 3) because of
[] dup 2dup 4dup 4dup 4dup ...