Number series

[baric]

This may be too easy, but does anyone know what the next number in this series would be

 

1, 2, 3, 7, 43, 13, 53, ?

[DrRocket]

This may be too easy, but does anyone know what the next number in this series would be

 

1, 2, 3, 7, 43, 13, 53, ?

 

This sort of question comes up from time to time, and often whoever wrote the question has some clever definite answer in mind. The plain truth is that there is no pat answer and the next number could be literally anything.

 

It is quite possible, and in principle it is simple, to generate a polynomial whose value at 1,2,...,n is any set of n numbers. That does not guarantee that the next element in a sequence is given by that polynomial at n+1.

[baric]

This sort of question comes up from time to time, and often whoever wrote the question has some clever definite answer in mind. The plain truth is that there is no pat answer and the next number could be literally anything.

 

It is quite possible, and in principle it is simple, to generate a polynomial whose value at 1,2,...,n is any set of n numbers. That does not guarantee that the next element in a sequence is given by that polynomial at n+1.

 

Well, I understand that a polynomial can be constructed to map any desired series, but I did have a particular number in mind!

 

It was intended as a puzzle, not as a request for a mathematical proof

 

 

The next value is 5.

 

The series represents the prime numbers generated by starting with 1 and using Euclid's proof of infinite primes to generate a list of prime numbers.

[timo]

Point apart that one is not a prime number, shouldn't the series be 1,2,3,7,43,1807,3263443, ... in that case? Why is there a 13 after the 43?

[DrRocket]

Point apart that one is not a prime number, shouldn't the series be 1,2,3,7,43,1807,3263443, ... in that case? Why is there a 13 after the 43?

 

no

 

Euclid's proof goes as follows: Let [math]x_1,...,x_n[/math] denote the first n primes. I claim that there is a larger prime. So let

 

[math]p=x_1x_2...x_n=1 [/math]

 

and

 

[math]q = p+1[/math]

 

Now [math]q[/math] is not divisible by any of the [math]x_i[/math] so either q is prime or it is divisible by a prime between p and q. In either case there is a prime larger than [math]x_n[/math]

 

The critical point is that one considers the product of ALL primes less than or equal to some given prime.

 

The next number after 47 would be the product of all primes less than or equal to 43, plus one. That number is a lot larger than 1807.

[baric]

Point apart that one is not a prime number, shouldn't the series be 1,2,3,7,43,1807,3263443, ... in that case? Why is there a 13 after the 43?

 

It's a series, and numeric series generally start at 1, which is non-prime by definition rather than proof.

 

13 comes after 43 because (1*2*3*7*43)+1 is equal to 1807, which is NOT a prime number. According to Euclid, this product+1 will either be a prime or factored by a prime not in the original set of primes. The first prime factor for that 1807 is 13.

 

The critical point is that one considers the product of ALL primes less than or equal to some given prime.

 

The next number after 47 would be the product of all primes less than or equal to 43, plus one. That number is a lot larger than 1807.

 

Actually, Euclid's proof does not require the list of primes in the original set to be comprehensive over a given range. It simply requires that they be prime for his proof to work.

 

In other words, his proof will work for any randomly-selected set of prime numbers.

Edited September 5, 2011 by baric

[DrRocket]

 

Actually, Euclid's proof does not require the list of primes in the original set to be comprehensive over a given range. It simply requires that they be prime for his proof to work.

 

In other words, his proof will work for any randomly-selected set of prime numbers.

 

Actually, no.

 

And what I presented is the proof usually attributed to Euclid -- see for instance An Introduction to the Theory of Numbers by Hardy and Wright.

 

The statement doesn't even make sense in the context of "any randomly-selected set of prime numbers".

 

Moreover a product of primes, plus 1 is not necessarily prime.

 

3x5 + 1 = 16 which is most certainly not prime even though 3 and 5 are both prime.

[timo]

The critical point is that one considers the product of ALL primes less than or equal to some given prime.

Makes sense. Thx for the comment.

[baric]

Actually, no.

 

And what I presented is the proof usually attributed to Euclid -- see for instance An Introduction to the Theory of Numbers by Hardy and Wright.

 

The statement doesn't even make sense in the context of "any randomly-selected set of prime numbers".

 

Moreover a product of primes, plus 1 is not necessarily prime.

 

3x5 + 1 = 16 which is most certainly not prime even though 3 and 5 are both prime.

 

 

yes, but 16 is factorable by 2, which is a prime not in your original set of primes. What Euclid's proof provide is to allow you to extend ANY set of randomly selected prime numbers by providing a new number that is either a prime not in the original set or is composed by primes not in the original set.

 

Which particular primes are in the original set is irrelevant.

 

With regards to 1 not being prime, I think that is now accepted by consensus. The issue of its primality is an ancillary point unrelated to the series I listed.

[DrRocket]

yes, but 16 is factorable by 2, which is a prime not in your original set of primes. What Euclid's proof provide is to allow you to extend ANY set of randomly selected prime numbers by providing a new number that is either a prime not in the original set or is composed by primes not in the original set.

 

Which particular primes are in the original set is irrelevant.

 

With regards to 1 not being prime, I think that is now accepted by consensus. The issue of its primality is an ancillary point unrelated to the series I listed.

 

 

All that you have said is that if you take a list of primes, multiply them and add 1 then the resulting number is not divisible by any of the primes in the list. That is true, but trivial. In contradiction to your assertion, you need even confine the list to prime numbers, just integers other than 1. This is simply because if n divides both p and p+1 it must also divide 1.

 

This is similar to the proof in my copy of The Elements, modulo recasting in less archaic language. This proof does not produce any identifiable sequence, as suggested by your OP, but merely shows that no finite list of primes can be complete. That is the important number theory result. But it has nothing to do with the production of your "series".

 

 

Short version: This discussion is now just plain silly.

[baric]

All that you have said is that if you take a list of primes, multiply them and add 1 then the resulting number is not divisible by any of the primes in the list. That is true, but trivial.

 

I never said it was non-trivial. It is simply the foundation of Euclid's proof, which is the algorithm I used to generate the numeric series. It was intended as a puzzle, not a deep mathematical insight.

 

In contradiction to your assertion, you need even confine the list to prime numbers, just integers other than 1. This is simply because if n divides both p and p+1 it must also divide 1.

 

I have never argued that 1 was prime. It was simply a useful starting point for the series, much the same way that 0 & 1 are simply defined to be the starting values of the Fibonacci sequence.

 

This is similar to the proof in my copy of The Elements, modulo recasting in less archaic language. This proof does not produce any identifiable sequence, as suggested by your OP, but merely shows that no finite list of primes can be complete. That is the important number theory result. But it has nothing to do with the production of your "series".

 

Euclid's proof does not provide an identifiable series. As you said, it is a trivial proof by today's standards. But the application of his proof to an initial set of primes (I chose to start with the unit 1, but 2 would have sufficed), does produce a series.... the one I listed in the OP as a simple puzzle.

 

 

Short version: This discussion is now just plain silly.

 

Really. I must have missed the transformation from pedantic.

Edited September 5, 2011 by baric