Problems & Puzzles: Puzzles

Puzzle 117. Certain p# +1 values

We know that p#+1 can not be divided by any prime from 2 to p; then p#+1 has a prime divisor greater than p. As a matter of fact this is exactly the Euclid's proof of the infiniteness of the quantity of prime numbers.

Question

If we ask for such numbers p#+1 divided precisely by the next larger prime to p, then the first 4 cases are: p = 2, 17, 1459, 2999. (See A058233) Can you find the next 3 cases?

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Note: For sure you know that p# is the "primorial of p" or the product of all the primes less than  or equal to p.
Observation: This puzzle mainly tests your skill to handle large numbers. For p=2999 the number p# +1 has 1274 digits.

Solution

Paul Jobling reports (December 5, 2000) that he found no one more example for 2999<p<=20,000,000.

Previously he explained to me that by no means this puzzle implies to handle very large numbers (as I naively supposed in the "observation") because p# can be calculated mod q (and then never greater than q) in a loop of n steps, where n is the index of the prime p, and q is "the following prime to p" of course.

Finally he sent to me the source and its corresponding executable of a code he made & used in his search, in order to be provided on request to any reader who wants to continue the search beyond the range he covered this week.

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In Ubasic the code to get the solutions asked in this puzzle is:

10 P=2:'puzz117.ub
20 P=nxtprm(P)
30 Q=1:F=1
40 if nxtprm(Q)=P then goto 60
50 Q=nxtprm(Q):F=(F*Q)@P:goto 40
60 if (F+1)@P=0 then print Q,P
70 goto 20

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Wilfred Whiteside wrote (November, 2005):

I decided to try my luck at running puzzle 117 on my new dual core.  It had been almost 5 years since Paul Jobling's result, so I hoped to expose it to increased fire power with the chance of finding something.

Paul Jobling reports (December 5, 2000) that he found no one more example for 2999<p<=20,000,000."

I searched up to p=272,112,959 (the 14,820,001th prime) with no solution.  I had the program print all remainders r= (P#+1) mod Pnext that were <=20.  Given how rare remainders less than or equal to 20 are (about 3 or 4 for every power of 10), I came to realise that the next number solving this puzzle is likely quite large.  Given that the compute power to test up to the nth prime goes like n^2, of course, makes matters far worse.  I stopped after a 2x3.2 GHz month.

I also looked for the case of a remainder r=2, which would mean a solution to (P#-1) mod Pnext = 0, a modified puzzle 117.  After 3 and 7, only 1 number in the range 7<p<=272,112,959 showed up as a solution!  The magic number was 176,078,267.

So, if puzzle 117 had been posed as find the next P after 3 and 7 where (P#-1) mod Pnext=0, the sequence of numbers would be 3, 7, 176078267  (assuming my code was working correctly)

Note:  Below are the numbers p and the remainders r=(P#+1) mod Pnext where the remainder was <=20, for p>100,000.  This showed me how hard a solution might to puz117 might be to find!

p=158,071 -> r=17
p=325,079 -> r=9
p=362,927 -> r=13
p=2,186,389 -> r=8
p=3,539,831 -> r=17
p=7,129,561 -> r=10
p=9,573,937 -> r=14
p=12,553,753 -> r=10
p=75,236,351 -> r=11
p=109,818,281 -> r=5
p=129,160,789 -> r=15
p=176,078,267 -> r=2
p=266,350,537 -> r=14

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