We define a floor exponent prime sequence (FEPS) to
be a sequence of primes of the form {p_{1}, p_{2}, ... ,
p_{r}}, where p_{i} =
Floor{t^{i}}
for some t>2. For example, for t=1287/545, we get the sequence {2, 5, 13,
31, 73, 173, 409, 967} (see **Crandall** and **Pomerance**, "Prime
Numbers, a Computational Perspective" (2000), exercise 1.75, p.69). **
Crandall** and **Pomerance** ask whether any longer such sequences
exist, whether such a sequence can be infinite, and whether an infinite
number of such sequences of length 3 exist. In 1989, **A. Balog**
showed that there are infinitely many such sequences of length 2.

Recently I began to study this problem. After a few
days, I discovered a rather simple mechanical procedure for generating all
FEPS with base t up to a given bound. Let FEPS(t) denote the floor
exponent prime sequence with base t and let EFEPS(t) denote the sequence
augmented by one (composite) element. Thus, FEPS(2) = {2} and EFEPS(2) =
{2,4}. We proceed as follows: Call a base t a boundary value if t is
FEPS(t1) is different from FEPS(t) for all t1<t. Suppose t1 is a boundary
value. Given a FEPS(t1) = {p_{1}, p_{2}, ...,
p_{r}, p_{(r+1)}} where
p_{i} =
Floor(t^{i}) is prime for 1<=i<=r and
composite for r+1, compute the next boundary value t2 by computing the
minimum of (p_i+1)^{1/i} and
NextPrime(p_(r+1))^{(1/(r+1))}. For example, when t1=2, we have
EFEPS(t1)={2,4}, so t2 = min(3,sqrt(5)) = sqrt(5) and
FEPS(sqrt(5))={2,5,11} is the next FEPS and EFEPS(sqrt(5))={2,5,11,25}.
(We get 11 for free here since 11 < t2^{3} = 5^(3/2)). We may continue in
this way to get all FEPS(t) in this way with t up to some bound X. I did
so during the past few days and was rather surprised by the results! It
didn't take long to discover the first FEPS of length 9, namely
FEPS(3450844193^{(1/9)})={11,131,1511,17341,
198997, 2283583, 26205133, 300715537, 3450844193} (Sloan's sequence
A076255) and the first of length 10, namely FEPS(39661481813^{(1/10)})={11,131,1511,17351,199151,2285711,26233621,301089179,3455668247,39661481813}

(Sloan's sequence A076357). Shortly thereafter I
discovered thousands more, most of them as long or longer than these! Thus
far, the longest such sequence I've found is:

FEPS(245092494303867872316927840433058150121090972563187226252192162731242781828213\

525601390456874200763428926486556401229467984897295568912536209575405704779365\

881331794192534581134284391993662578240712109237575614929981232768530259462270\

2807975786215971789524086870912809903356207088970031733058733884867^{(1/100)}),

which has length 100.

Based on numerical analysis of the FEPS I've
produced so far, I propose the following conjectures:

**1. For a fixed positive
integer k, let π**_{k(x)} denote the number of FEPS with base t<=x. Then
π_{k(x)} is asymptotic to 1/k! (x/log(x))^{k} as x approaches infinity.
Note that for k=1, this reduces to the prime number theorem.

**2. There are infinitely
many FEPS of length k for all k>1.**

**3. There are no infinite
FEPS.**

**4. Let
Π(x) denote the
total number of FEPS with t<x. Then log(Π(x)) is asymptotic to x/log(x)
as x approaches infinity.**

**5. For fixed x, the
distribution of lengths of FEPS is asymptotically Poissonian with mean
x/log(x) as x approaches infinity.**