## 12 Reputation

18 years, 172 days

## Many thanks. The program has...

Many thanks. The program has proved incredibly helpful; I was able to find a counterexample to a conjecture that had looked quite promising! How shall I credit you in my report? Thanks Ben

## Many thanks. The program has...

Many thanks. The program has proved incredibly helpful; I was able to find a counterexample to a conjecture that had looked quite promising! How shall I credit you in my report? Thanks Ben

## Hi Joe, Many thanks for your...

Hi Joe, Many thanks for your reply again. Your assumption about how the mod 2 expession applied was quite right. The algorithm didn't seem to work as I had calculated. I may be quite wrong but I couldn't spot the condition epsilon(p+1)=epsilon(p) (possibly not helped by a slightly illogical ordering of initial values on my part); I wonder if that's the reason why. If it's not too much hassle would you be able to insert it? Here's the whole algorithm again in hopefully the clearest form yet. Initial values: epsilon(i) = i mod 2 if i is less than or equal to p epsilon(p+1)= epsilon(p) alpha (1) = 0 alpha (2) = 1 Recursive relationships epsilon(i) = [epsilon(i-p) - epsilon(i-1) - alpha(i)] (mod 2) for i greater than or equal to p+2 alpha (i)= 0 if alpha(i-1) + epsilon (i-1) + epsilon(i-2) is less than or equal to 1 and = 1 if alpha(i-1) + epsi (i-1) +epsi (i-2) is greater than 1 Many thanks again, Ben

## Hi Joe, Many thanks for your...

Hi Joe, Many thanks for your reply again. Your assumption about how the mod 2 expession applied was quite right. The algorithm didn't seem to work as I had calculated. I may be quite wrong but I couldn't spot the condition epsilon(p+1)=epsilon(p) (possibly not helped by a slightly illogical ordering of initial values on my part); I wonder if that's the reason why. If it's not too much hassle would you be able to insert it? Here's the whole algorithm again in hopefully the clearest form yet. Initial values: epsilon(i) = i mod 2 if i is less than or equal to p epsilon(p+1)= epsilon(p) alpha (1) = 0 alpha (2) = 1 Recursive relationships epsilon(i) = [epsilon(i-p) - epsilon(i-1) - alpha(i)] (mod 2) for i greater than or equal to p+2 alpha (i)= 0 if alpha(i-1) + epsilon (i-1) + epsilon(i-2) is less than or equal to 1 and = 1 if alpha(i-1) + epsi (i-1) +epsi (i-2) is greater than 1 Many thanks again, Ben

## Corrected and simplified algorithm...

Thanks for your reply; it’s really appreciated. I realised I made some stupid mistakes in my last post and over-complicated things. Yes ‘p’ is a constant which you pick at the start. So here we go; The algorithm consists of 2 interacting elements; alpha(i) and epsi(i). The object is to compute the first however many (say 1000) iterations of epsi(i). To do this you need the first however many iterations of alpha(i) Could you give me an indication of how difficult it would be to write a program. (Or write one and e-mail it to me if there is a very kind soul out there who would enjoy this kind of thing.) I'm a complete novice and I want to use the algorithm to test a conjecture relating to the collatz conjecture for my undergraduate maths project. Many, many thanks. Here it is: Initial values: eps(i) = 0 if i=2k where k is a natural number less than or equal to p eps(i) = 1 if i=2k-1 where k is a natural number and i is less than or equal to p alpha (1) = 0 alpha (2) = 1 epsi(p+1)= epsi(p) Recursive relationships epsi(i) = epsi(i-p) - epsi(i-1) - alpha(i) (mod 2) for 1 greater than or equal to p+2 alpha (i)= 0 if alpha(i-1) + epsi (i-1) +epsi (i-2) is less than or equal to 1 and = 1 if alpha(i-1) + epsi (i-1) +epsi (i-2) is greater than 1

## Corrected and simplified algorithm...

Thanks for your reply; it’s really appreciated. I realised I made some stupid mistakes in my last post and over-complicated things. Yes ‘p’ is a constant which you pick at the start. So here we go; The algorithm consists of 2 interacting elements; alpha(i) and epsi(i). The object is to compute the first however many (say 1000) iterations of epsi(i). To do this you need the first however many iterations of alpha(i) Could you give me an indication of how difficult it would be to write a program. (Or write one and e-mail it to me if there is a very kind soul out there who would enjoy this kind of thing.) I'm a complete novice and I want to use the algorithm to test a conjecture relating to the collatz conjecture for my undergraduate maths project. Many, many thanks. Here it is: Initial values: eps(i) = 0 if i=2k where k is a natural number less than or equal to p eps(i) = 1 if i=2k-1 where k is a natural number and i is less than or equal to p alpha (1) = 0 alpha (2) = 1 epsi(p+1)= epsi(p) Recursive relationships epsi(i) = epsi(i-p) - epsi(i-1) - alpha(i) (mod 2) for 1 greater than or equal to p+2 alpha (i)= 0 if alpha(i-1) + epsi (i-1) +epsi (i-2) is less than or equal to 1 and = 1 if alpha(i-1) + epsi (i-1) +epsi (i-2) is greater than 1
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