Problems & Puzzles:
Puzzles
Puzzle 227. Research
Problem 1.75
In the Research Problem 1.75 of the
Crandall's and Pomerance book "Prime numbers, a computational
perspective", p. 69 we may read:
"for t=1287/545, the integer part of
the first 8 powers are 2,5,13,31,73,173,409,967, each of which is prime.
Find a longer chain"
It's obvious that the integer part of
the 9th power (2283) must be composite.
For this puzzle we will interpret
this "find a longer chain" in the following way:
Q.1. Find another
t value such that the integer part of the first 9 or more powers are
primes.
An additional question is this one:
Q.2 Can you
explain why Crandal & Pomerance (and others) used t as
1287/545
and not 111/47
or many other equivalentforthepurpose fractions?
_____
See Problem 42,
posed by David Terr
Solution:
Contributions cam from Adam
Stinchcombe, Luke Pebody & Giovanni Resta.
Stinchcombe wrote:
You can generate long list of such
values as follows: Choose any prime p1, then t will be between p1 and
1+p1, square this inequality to get t^2 between p1^2 and (1+p1)^2, then
list all the primes in this range. Say a sample prime here is p2. Then
t^2 is between p2 and 1+p2, take this inequality to the 3/2 power to get
an inequality for t^3, list primes in that range, take those primes to the
4/3, etc. Eventually this process will exhaust itself  the range for
t^(high power) will contain no primes for t within the bounds previously
established. Then take a rational approximation, e.g., continued fraction
convergent, to the root of the final prime in the sequence and use this
for t.
The value t=227320025/12874973
produces the primes floor(t^j)= 17, 311, 5503, 97177, 1715761, 30293413,
534859331, 9443455657, 166733287613, 2943836473181, 51976262837953,
917690885078381 for the values j=1..12.
As a "proof of concept," the prime
p=1009 yielded the value
t=395497033627864864932979264416357624654833134867/391624669740679761871064122844303957704237961
which produced 30 primes (actually, 30 prp's, using Maple's isprime
feature). They are: 1009, 1019873, 1029958121, 1040142292613,
1050427164233453, 1060813732116394603,...,
1343370060330352532785078287202759129080447511955678414038366444593248871378411352356866591
***
Pebody wrote:
2077193/181012 produces 9
5475543/477079 produces 10
18973550/1440073 produces 11
258868631/18620263 produces 12
1678112765/95690094 produces 13
12341592517/703747789 produces 14
22093296417/1259813795 produces 15
81211370385/4630870957 produces 16
***
Resta wrote:
...here are some solutions for powers from 9 to 20
for puzzle 227
powers 1..9 t = 941603 / 82041
primes =
11, 131, 1511, 17351, 199151, 2285711, 26233619,
301089149, 3455667857
powers 1..10 t = 5475543 / 477079
primes = 11, 131, 1511, 17351, 199151, 2285711,
26233621, 301089179, 3455668247, 39661481813
powers 1..11 t = 18973550 / 1440073
primes = 13, 173, 2287, 30133, 397027, 5230997,
68920531, 908056153, 11964010741, 157630728439, 2076849234433
...
powers 1..20 t = 9245545434233933 / 315724048003569
primes = 29, 857, 25111, 735359, 21533983,
630593149, 18466054931, 540753075169, 15835211656507, 463712440521139,
13579182404278663, 397647720128968919, 11644567737299383501,
340995184744365951731, 9985575990630243623701, 292413888305789764542331,
8562939399223795207216823, 250753928206382544084651491,
7342984643407766159814138311, 215029227494071397857756115239
Using my approach it will be very difficult to find
a value for 50 primes and impossible to find values for 100 primes.
My actual record (and I stopped my search there) is
35 powers for
t=72013968170625248106344150215148/1334984596802709772861985688627 where
the last prime, that is floor(t^(35)) has 61 digits.
My approach is the following:
I wrote a recursive Mathematica function which,
given a power p and a floating point range (a,b), find which subranges
(aa,bb) of (a,b) such that floor(aa^p)=floor(bb^p)=a prime. Then it calls
itself in the range (aa,bb) for power p+1. More or less. It then records
the maximal p reached in the given starting interval. At the end I obtain
a interval which is not further divisible.
It has normally the form q^(1/p) ... (q+1)^(1/p)
where q is the largest prime I will obtain as floor(t^p).
At that point I use the Rationalize function of
Mathematica to find a rational number (possibly with a relatively small
denominator and numerator) which is inside the interval (I ask for a
rational which approximate the middle of the interval with an
approximation of half the interval span), so I obtain a rational t.
The maximal power reachable grows (more or less)
with the prime interval considered. For example I reached 20 (22 really)
in the interval (29<t<30) and 35 in the interval (53<t<54).
I estimated that 50 will be reached for t>75 and 100
for t>140. This means that the last primes obtained will be around
(75)^(50) and (140)^(100) which are pretty big numbers and require a lot
of precision digits during the floating point calculations.
Moreover the recursive exploration of an interval
(at least with my simple method) require a number of recursive calls with
grows more or less exponentially with the interval considered. Say, they
are 259098 calls for (47<t<48) and 595066 calls for (53<t<54) where
Mathematica spent several hours. So I think that with this approach I can
reach 50 only recoding the algorithm in C using a multiple precision
library (but I do not have interest in it, actually), while the case of
100 primes is completely beyond this approach.
***
Phil Carmody wrote (12/03/04):
For puzzle 227, I've found some smallest sequences of length 36 and 37.
8327848885667459649905801739208369196325130888274609500504483297^(1/36)
8327848885667459649905801739208369196325130888274609500504483337^(1/36)
429770745544112650251740197448446451738502486973260949027731341197^(1/37)
569831552238384205382549706997729158203052589452328929163635282383^(1/37)
The smaller 37 has simplest continued fraction approximant
415362374334426054152330893235404659/6991266980150161592067385870604371
As the initial prime range, [5960) in the above, increases, I'd expect
to see many more longer sequences. I believe that the starting point of
[10091010) can be bettered for a length 100 sequence.
I'm not sure I agree with David Terr's conjecture 5, as it seems to
imply that for a starting point of 1009, the average length should be 145,
in which case David's length 100 one was a particularly poor sequence,
which makes no sense to give as an example. However,
that might be the case, and the conjecture might be sound. I'll have to do
some more investigating.
***
