Problems & Puzzles:
Puzzles
Puzzle 164. Sequences of Fermatian primes
The following puzzle has been posted by Daniel Gronau.
It's about the Fermat Numbers, usually
described by the function: F_{n} = 2^(2^n)+1, for n=0, 1, 2, .... Gronau
wrote:
"There is an alternative description  a recursive one  of Fermat numbers:
f(n+1) = 2 + f(0)*f(1)*f(2)*...*f(n).
f(0) = 3
Following this "rule" we get the five (known) Fermat
successive numbers & primes:
3, 5, 17, 257, 65537.
So I asked myself if we can get again five successive primes
if we substitute f(0) with a prime other than 3.
The answer was: YES! As a matter of fact I found five successive primes
solutions for the following f(0) values: 19379, 9443669, 11054759, 15992069, 22482329,
32557379, 51102509, ..."
Daniel wondered if the quantity of successive primes obtained by
this recursive procedure were limited or not.
I (C.R.) found several f(0) values producing six successive primes.
Here are the first 3 of them:
1) 1288004999
2) 1417870889
3) 1619225519
Question:
1. Can you find 3 more f(0) solution
for sequences of six
successive Fermatian primes?
2. Can you find f(0) solutions
for sequences with seven
and eight
successive Fermatian primes
Solution:
Sudipta Das notices that the
recursive expression of the Fermatian numbers, given by Gronau:
f(n+1) = 2 + f(0)*f(1)*f(2)*...*f(n).
f(0) = 3
Can be transformed in a closed
expression:
f(n) =
(f(0)+1)^(2^(n1)) + 1, for n>0
f(0) = 3
Sudipta claims that this expression
"is much easier to program and to test for primality, than the
recursive form"
***
Shyam Sunder Gupta has found 3 more
f(0) prime values producing sequences of six consecutive primes:
3998094509
4997719499
7369307219
***
Both Shyam and Sudipta point
out that f(0) must leave a remainder of 29 when divided by 30.
***
Jim Fougeron made an extensive search and here
are his results:
I have found no "order 7" or "order 8". I have
search ALL ranges up to > 1.6767 trillion. Within this range, there
were 217 numbers which satisfied 6 consecutive primes, and 13927 numbers
which satisfied 5 consecutive primes (not counting the 217 order 6
numbers). I tested some of the 7th candidate numbers (factoring), and it
appears that there is no "pattern" which would eliminate a 7th
number being prime (i.e. all were not divisible by 3 or 13 > or 101,
...). It also appeared that the 7th number's behave as to be partly
factored, similar to the Fermat numbers, the General Fermat numbers, or
the Mersenne type numbers. The prime density would certainly be more
than simply a "random" 300 to 400 digit number. At 200+ order
6's I did "expect" to see an order 7 result, but none were
found :((((
I did also observe a few "rules" for this sequence.
1. The f(0) and f(1) must be a twin, since f(1) = f(0)+2
2. f(0)%10 must be 9 : (except for f(0)=3)
if f(0)%10 is even, then f(0) is not a prime
if f(0)%10 is 1, then f(10)%10=3 and 3*1+2 is 5, so f(2) is divisible by
5
if f(0)%10 is 3 then f(1)%10=5 so it is not prime (except if f(0)=3)
if f(0)%10 is 5 then f(0) is not prime (except if f(0) is 5)
if f(0)%10 is 7 then f(1)%10 is 9 and 7*9+2 is 65%10=5.
if f(0)%10 is 9 then f(1)%10 is 1 and f(2)%10 is 9 and f(3,4,...)%10
is 1
3. f(n) as n grows it will end with many 0's followed by a single 1
digit, so MANY (most) of these primes, even when large, will be VERY
easily provable (N1 will be 10^many*something with 10^many often being
larger than N^1/3 which allows easy N1 proving)
***
I asked to Jim: "Can you also add the type of program (language),
computer and time spent in your search? One more question: trillion is
like in our system 10^18?". And his answer was:
1 trillion is 2^12 (1,000,000,000,000). I am currently at
3.845*10^12 and
have found 187 more order 6 values, but no order 7's yet.
I have coded this in VC++ using my own 64 bit Erat primegenerator to
find
the twins, and then Miracl large integer library to prp test the
"larger"
numbers. Coding took a few hours, and runtime to 10^12 took 3
Athlon 750mhz
PC about 16 hours each. I have optimized my code better, and now a
single
PC running for about 30 hours pushed from 1.2*10^12 to 3.8*10^12.
Note that
the size of the f(4+) is increasing now, so testing the numbers is
slowing
down. I do expect a series of length 7, but not sure how high I
will have
to push to find it. I will not be searching for a length 8 item ;)
I will
leave that to someone else. I currently have over 400 length 6
results found.
Here is some new information. [higher modular requirements for
f(0)]
The f(0) numbers in the series (other than f(0)==3) MUST be of the form
f(0)==509 mod(510) That is why 9 mod(10) and 29
mod(30) were both "valid".
510 is 2*3*5*17 so any number which is made up exclusively of these
factors
(10=2*5, 30=3*2*5, 85=17*5, ...) would all "satisfy" the f(0)
to be n1 mod(n).
However, 509 mod(510) is the optimal (maximal) case.
***
A few hours later Jim informed a beautiful result:
Carlos, here is an update on puzzle 164
I have processed from f(0)=3 to f(0)=4319420014079 and here are the
results:
There were 29777 length 5 finds
There were 456 length 6 finds
There is 1 length 7 find. That find is:
f(0)=4109355036959
f(1)=4109355036961
f(2)=16886798819788522966041601
f(3)=285163974380011052145033104437715842266772930560001
f(4)=
813184922842035997316061285984072393325394068447975735939012051758969
56396923659519570581913600000001
f(5)=
661269718737608038306331633156409174030623576506013895765168627135044
165565655941249372213421363090241527339594205302044863667773950048188
0593852970658833909784236964257314836927437864960000000000000001
f(6)=
437277640919315243514787851517751082721195257036696929098144568407957
920797032668040071454163950082962790256662644321871244216898947128310
396937497654368156594418147998528995865975591284432223136928623664627
293496717205649428628521123863684172634387233026153590991365834794438
361887012729368899010291270461836799147284907986146416473268907016127
38703984886086481563195801600000000000000000000000000000001
Note f(6) is a 402 digit prime. Also note that f(0) is the smallest
value
which is prime for length 7. All primes are proven.
Jim.
What a splendid work!
***
The 24/3/2002 Phil Carmody came with an example
for length 8: f(0)+1=7,190,597,297,273,100
"... something like 3 weeks on my PPro200, and than a final
burst on my Duron 900 for a week finally yielded >550 length 7
results before providing a single solitary length 8 result:
f(0)+1 = 7,190,597,297,273,100
...
The filters required for this test are quite unusual, compared with
the usual tasks I've thrown at GenSv, and so I've had to make some
tweaks to the code to improve the efficiency in the sieve in these kinds
of situations. The thing I'm most proud of is the fact that the largest
term is titanic!
***
