Problems & Puzzles: Puzzles

 Puzzle 484. ABC conjecture and consecutive primes Anton Vrba posed this puzzle:  Based on the concept of the ABC conjecture we define PAB triples as P+A=B  with the condition that P is prime and the prime factors of the product A.B are consecutive primes. The quality of the triple is defined as q=Log(B)/Log(p# P) where p is the largest prime factor of A.B. Example:  Using the first 8 eight primes B = 19^2.17^3.11^22 A = 2^2.3^4.5^17.7^6.13^11 P = B - A =  is a 25 digit prime  and Log(B)/Log(19#  P) = 1.00086 Q1: Find the minimal solutions using the first 2, 3, 4, ... n primes. (P minimum and q>1)

Contributions came from Farideh Firoozbakht & Enoch Haga

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Farideh wrote:

n = 2: 2*3
P = 2, B = 2*3^2 = 18, A = 2^4 = 16 & q = Log(18) / Log(2*3 * 2) = 1.163

n = 3: 2*3*5
P = 2, B = 2*3^4 = 162, A = 2^5*5 = 160 & q = Log(162) / Log(2*3*5 * 2) = 1.243

n = 4: 2*3*5*7
P = 2, B = 2*3^2*5^2 = 450, A = 2^6*7 = 448 & q = Log(450)/Log(2*3*5*7 * 2) = 1.11

n = 5: 2*3*5*7*11 P = 2, B = 2*5^2*11^2 = 6050,
A = 2^5*3^3*7 = 6048 &q = Log( 6050 ) / Log( 2*3*5*7*11 * 2 ) = 1.032

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Enoch twisted a little bit the puzzle and sent:

My interpretation of this puzzle may be a little different from what was
intended by the example in which prime factors are raised to powers and where A is the greatest or last prime factor in the sequence.

My approach is to take the sequence of primes q: 2 3 5 7 11 13 17 19 . . . p, and break the sequence into two parts where the product of factors in each is as equal to the other as possible. In practice we must have P=B-A where A is
smaller. The goal is to simultaneously produce P as small as possible and q
as near to 1 as possible, where the ratio is >1. My approach is to
keep the sequence "clean" by not raising factors to powers in an effort to
produce a lower value for P.

In some cases I note that it may be valuable to allow P=1, thus:

2 factors: 2*3
P=1 B=3 A=2 q=log(3)/log(2)=1.5850

3 factors: 2*3*5=15*2=6*5
P=13 B=15 A=2 q=log(15)/log(2)=3.9069 or
P=1 and log(6)/log(5)=1.1133

4 factors: 2*3*5*7=35*6=15*14
P=29 B=35 A=6 q=log(35)/log(6)=1.9843, or
P=1 and log(15)/log(14)=1.0261

5 factors: 2*3*5*7*11=55*42
P=13 B=55 A=42 q=log(55)/log(42)=1.0721

6 factors: 2*3*5*7*11*13=210*143
P=67 B=210 A=143 q=log(210)/log(143)=1.0774

7 factors: 2*3*5*7*11*13*17=935*546
P=389 B=935 A=546 q=log(935)/log(546)=1.0853

8 factors: 2*3*5*7*11*13*17*19=3230*3003
P=227 B=3230 A=3003 q=log(3230/log(3003)=1.0091

I theorize that log(B)/log(A) approaches 1 at infinity.

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