Problems & Puzzles: Puzzles

Puzzle 320. Dealing with primorials Rudolph Knjzek sent the following puzzle: There is an idea for a new puzzle dealing with the primorial p#: Let p(k) be the k-th prime and p#(k) the product of the first k primes. Define a(k) = p#(k)+1 Define b(k) = p#(k)+q ; q an odd prime Define c(0) = 1 c(k) = p#(k+1)+c(k-1) for k>0 The a(k) are primes or composites of primes greater than p(k). Conjecture 1: The sequence a(k) contains an infinity of primes. If p(k)>=q q divides b(k). Conjecture 2: The sequence b(k) contains at least one prime for all odd primes q. Assertion: The sequence c(k) contains only a finite number of primes. Questions: 1.Who can prove or disprove Conjecture 1? (seems to be impossible) 2.Who can prove or disprove Conjecture 2? (maybe there is a counterexample) 3.Who can prove the above assertion about the sequence c(k)? (easy)

 

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Contributions came from Adam Stinchcombe, Faride Firoozbakht and J. K. Andersen:

Adam wrote:

Conjecture 2, I have listed below record "waits," the longest time you have to wait to make bk a prime.  The list is the value q and the value pk. so p#(j)+q is not prime for pj<pk:

 

7,3

19,7

89,11

683,13

929,17

1471,19

1559,43

10781,107

21019,241

76519,251

78041,383

91621,449

343411,457

601319,463

664777,499

702991,2069

 

My computer got stuck for larger values.  It seems that for large q, there will be enough candidate p#(k) so that one bk would have to be prime.  I found no counterexample through q=702991

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Farideh wrote:

1. Conjecture 1 says the sequence A014545(1,2,3,4,5,11,75,171, 172,384,457,616,643,1391,1613,2122,2647,2673,4413,13494,31260,
33237,...) is infinite.
I think if the conjecture 1 is true the proof is very difficult or it is impossible to prove.

2. I think for conjecture 2 we can not find a counter example and in fact there is no such example. But I don't know how we can prove this conjecture. I checked it for the first 1615910 primes.

3. It is obvious that if t<n+2 & p=prime(t) divides c(n-1) then for each j>=n-1 p divides c(j).
According to the above statement since 763=prime(90) divides
c(88) so for each j>=88, 763 divides c(j). Hence The sequence
c(k) contains only a finite number of primes.

All prime terms of this sequence are:

c(1),c(2),c(4),c(5),c(6),c(7),c(10),c(13),c(20),c(40),
c(41),c(42),c(73),c(77).

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J. K. Andersen wrote:

I use the normal notation p# for the product of all primes <= p.

1. Heuristics support an infinite number of p#+1 primes, but nobody
knows how to prove it.

The known primes are at http://primes.utm.edu/top20/page.php?id=5 :
p = 2, 3, 5, 7, 11, 31, 379, 1019, 1021, 2657, 3229, 4547, 4787,
11549, 13649, 18523, 23801, 24029, 42209, 145823, 366439, 392113.


2. I have less confidence in this conjecture.
It holds for q<10^8 with prp's (probable primes) allowed.

Below, p# + q is the first prime (prp's allowed) for that odd prime q.
The list of q which need a larger p than all smaller q:

p# + q
------
2# + 3
3# + 7
7# + 19
11# + 89
13# + 683
17# + 929
19# + 1471
43# + 1559
107# + 10781
241# + 21019
251# + 76519
383# + 78041
449# + 91621
457# + 343411
463# + 601319
499# + 664777
2069# + 702991
4783# + 2173663
8273# + 9069217
9767# + 19582841
45979# + 25852909

The GMP library and PrimeForm/GW prp'ed and found a prp for all q<10^8.
Only 45979#+25852909, 14923#+48265883 and 11321#+77045201 needed p>10000.

q=103476253 is the first where I have no prp. It has been prp'ed to 30000#+q.
It is infeasible to show that a q value above 10^8 is a counter example,
if the proof method is exhaustive prp testing.
(10^8)# is around e^(10^8) with around 43 million digits.


3. c(k) is prime for k = 1, 2, 4, 5, 6, 7, 10, 13, 20, 40, 41, 42, 73, 77.
p(90)=463 divides c(88). This means 463 divides c(k) for all k>=88.
c(k)/463 is prp for k = 96, 131, 135, 138, 168, 172, 462, 1006.

p(5704)=56237 divides c(5702), so 56237 divides c(k) for k>=5702.
463 and 56237 are the only fixed prime factors below 5 million.
The sum over primes of 1/p diverges, so I guess infinitely many primes will
become fixed factors. I also guess dividing by the fixed factors will give
infinitely many primes (similar to the p#+1 conjecture).

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