Problems & Puzzles: Puzzles

Puzzle 654 Factors for p(n)#+/-1

In one of the the Prime Curios! pages from Caldwell & Honaker, Jr. I found this curio:

510511 produces the first number of the form p(n)#+1 to have a prime factor of p(n+1), where p(n) and p(n+1) are consecutive primes.


Q1 Redo this for p(n) & p(n-1)

Q2 Redo both things now for p(n)#-1
(in both cases, send your largest finding)

Contributions cam frome Hakan Summakoğlu, Emmanuel Vantieghem, Jan van Delden & J. K. Andersen


Hakam wrote:

For p(n) and p(n+1):

p(n)#+1: (For i=1 to 232 product ( prime[i]))+1=613 digits number and have prime[233].

p(n)#-1: 5=2*3-1=5


For p(n) and p(n-1):

There isn't any number with these conditions.Because,

p(n)#+1: (For i=1 to n product ( prime[i])) +1 mod prime[n-1]=1

p(n)#-1: (For i=1 to n product ( prime[i])) -1 mod prime[n-1]=-1=prime[n-1]-1


Emmanuel wrote:

Obviously,  p(n)# + or - 1  cannot be divisible by  p(n-1).
My largest  n  such that  p(n)# + 1  is divisible by  p(n+1)  is  430 (with  p(n) = 2999,  p(n+1) = 3001.  There is no bigger such  n below 100000.
My largest  n  such that  p(n)# - 1  is divisible by  p(n+1)  is  4.  There is no bigger such  n  below  100000.


Jan van Delden wrote:

I checked untill p[170000]#+/-1, having more than 10^6 digits.
Type p(n)#+1, n=7   p(n+1)=19
Type p(n)#+1, n=232 p(n+1)=1471
Type p(n)#+1, n=430 p(n+1)=3001
Type p(n)#-1, n=2 p(n+1)=5
Type p(n)#-1, n=4 p(n+1)=11
Since p(n)#+/-1=+/-1 mod q if q in [2..p(n)] there can not be solutions with q=p(n-1).


J. K. Andersen wrote:

A prime has itself as prime factor so the first case is technically 2#+1 = 3.
Puzzle 117 asked for p(n)#+1 to be divisible by p(n+1).
The only known cases are p(n) = 2, 17, 1459, 2999 (A058233)

If I'm interpreting the question correctly then there is clearly no solution.
p(n)# is divisible by p(n-1) so p(n)#+1 can never be.

p(n)#-1 is divisible by p(n+1) for p(n) = 3, 7, and no others below 10^7.
Like in Q1, p(n)# is divisible by p(n-1) so p(n)#-1 can never be.



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