David Sycamore

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6 years, 12 days

MaplePrimes Activity


These are replies submitted by David Sycamore

@Carl Love Hi Carl,

Thank you for your very comprehensive response to my question, I hope you found it as interesting to do as I did to receive. I agree your analysis of the zeroes above,and was pleasantly surprised to see that you had extended the work to include numbers whose smallest prime divisor >3, something I had not anticipated.  

Best wishes

david  

@tomleslie Hi Tom, My Maple issue is 2017. Thanks for resending me the code. I followed your instructions and  was not asked about the application. I downloaded it directly to a Maple worksheet and tried to run it from the point you indicated. It did not work and returned the error message:


    "Error, (in CodeTools:-Usage) unable to execute seq". 

I am very grateful for all your assistance here, and all the insights and suggestions provided by Carl, which i have understood. In fact I had reached the same conclusion about proving the zeroes using congruences, eg since (3,5,7) correspond to (0,1,2) (mod 3), namely every equivalence class (mod 3) is there, then it follows that adding even k to all of them will result in one of those numbers being divisible by 3. Therefore any number which included 3,5 and 7 amongst its prime factors will always return a provable zero.The same approach works for all the other starting ponts I have tried ((3,5,13),(3,11,13), (3,7,23) and so on. However what I had not expected was to find zeroes for cases when the smallest prime is not 3 but 5,7,....., 37). This is new and very interesting and I had no trouble getting his code to run. As for running this one, maybe the eror message will enable you to understand why this happened, if so please let me know. Finally I am not familiar with all details of your code, example what does +~e mean ? If you know of a good textbook on Maple please let me know, my interest is mainly number theory as you can see. 

Best wishes

David. 


 

Hi Tom,

Thanks for this quick response. I am very interested in your results, had already seen the prime factor 3 thing and think perhaps it could be explained somehow (I don’t think it’s coincidence).  First I want to run the code for myself, get the data out and check with my hand cranking results then do some analysis. First problem is getting it to run. I am a novice with Maple so need the guide for dummies. Can you tell me what Instructions I need to add to your code to get it to run? Do I have to call the proc somehow? At the moment what I have done is copy your code (down to the end of the doWork proc ) into a worksheet and hit return. Then what I get is “unable to parse” . Obviously I am doing something wrong here. 

 

Questions: You define Nodd and Neven but they never seem to be called later in the script. However the doWork proc includes “N” , which N is this? I don’t follow, sorry. 

Regards

David

Many thanks for the response. Please advise how to modify the code so as to identify odd squarefree numbers n having more than 2 prime factors and the property p-1|n-1 only for the least and greatest prime divisors p of n. I have tried without success to do this myself. Assistance much appreciated.

David.

t1@Kitonum 

@vv Would it be possible to modify the previous code so as to output odd squarefree composite numbers having at least 3 prime divisors and the property that the only prime divisors p such that p-1|n-1 are the smallest and the greatest prime factors of n? 

The (provisional) data I have so far is: 231,1045, 1653, 4371, 4641, 5365, 6545, 8029, 9361, 10011, 10857

Best Regards

David Sycamore.

 

 

@Carl Love 

Regarding your comment about the squarefree function not being a check for squarefree but for returning the squarefree factorisation, I find this not to be quite the case.

Using with(NumberTheory) I find that IsSquareFree(n) returns the correct answer. Perhaps the capitalisation calls a different version?

Any comment?

@Carl Love Thanks,

By the way I found a way to compute all conforming terms up to an arbitrary N, just put your proc into a do loop; easier than I expected. 

@Carl Love 

I had not seen the (Kitonem)  comment about "squarefreee" before I replied to you last. Should something be changed ?

 

@Kitonum 

Thanks for your reply, which works for individual n as your example shows. How does the check for "squarefree" work? I thought there was a routine called "Squarefree" which could be called to do this but I dont see it in either reply. I woud like to adapt the code to compute all n<= any chosed N which have the property; is that possible?

Regards

David Sycamore.

@Carl Love 

Thanks Carl, 

It works fine for any selected individual number n. Could you suggest how I could adapt it so as to compute for some upper limit N, all n <= N which have this property?
Best regards

David.

@Kitonum 

Thank you very much, it works very nicely. I wonder if there is a modification you could suggest which would display the output in the form n, a(n) ? (so I can the correspondence see without counting every time)

Ceers

David.

@Carl Love 

Hi Carl,

Thank you.

To answer your question:

a(1)=1.

a(1)+1=2, a prime, so a(2)=a(1)+1=2

a(2)+1=3, a prime, so a(3)=a(2)+1=3

a(3)+1=4, composite and a(3) is odd so a(4)=a(3)+3=6

a(4)+1 =7, prime, so a(5)=a(4)+1=7.

a(6)=10

a(7)=11

a(8)=14, etc

Do you see what I’m trying to do?

Maybe my original explanation was not clear, hope this is better

David.

@David Sycamore 

Hello again. I ran into a problem with p=397 because the code runs out to N without finding a value for k,  and increasing N leads to a seemingly endless run. According to my calculation this prime should have a k value, and it should be a number == 0 mod 6. So I was wondering if the code could be adapted to restrict k >0 to values == 0 mod 6? This should mean faster computation, and if my hunch is right, produce a result. Would you mind to suggest something?

Thanks

David. 

@Kitonum 

Magic! 

Kitonum,

Thank you so much, you made my day. I need to learn how a proc works. Your code is a good starting point :-)

David. 

@Kitonum 

Thank you! I was glad to see k values for  79 and 167, as suspected. 

Can you please tell me where to enter the chosen prime into the code? I tried replacing p with 59 in the process, but this gave an error (invalid procedure parameters). Clearly I am doing something wrong.

Thanks

David.

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