David Sycamore

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1 years, 13 days

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These are questions asked by David Sycamore

As is well known, a number n is a Charmichael number if and only if for every prime factor p of n, p-1 divides n-1. 

I would like to find a way to identify the following: Odd square free composite numbers n, having at least one prime factor p, with the property that p-1 divides n-1.

obviously the Carmichael numbers are a sub sequence of this. I have already managed to write a code to identify odd square free numbers divisible by the sum of their prime divisors and am interested to see how these data will differ from those of  this new sequence. 

My problem here is to find, given an odd square free number, a way to select and test each prime divisor for the above divisibility requirement. I hope someone can help, thanks in advance

Best regards

David Sycamore.

Hi,

I am trying to write a code for the following simple recurrence:

a(1)=1,

a(n)+1prime—>a(n+1)=a(n)+1,

a(n)+1 composite —>a(n+1)=n+2

if a(n) even, or a(n)+ 3 if a(n) odd.

Data: 1,2,3,6,7,10,11,14,16,17.....

My first attempt is the following:

N:=10:

for k from 1 to N do

X:=1;

if isprime(X+1) then print(X+1);

elif not isprime(X+1) and mod(X,2)=0 

then print(X+2);

else print(X+3);

end if:

end do:

This does not work but I cannot see why. Would somebody mind to help me out with this?

 

Best regards

David.

 

 

 

I am trying to write a code to calculate, for a given prime p the string length k of the repunit

R_k=1111111....1111 (with k 1s) such that the number R_kpR_k is prime.  

Example: If p=59, then k=42 is the smallest k such that R_k59R_k is prime. Note that for some primes it can be shown that no such k exists (2,11,37,101..)

I can usually figure out from the factor cycle of a “repunit wrapped” prime whether such a k exists or not. But if the indications are that a k does exist then finding it is very time consuming. Eg for p =71, 167. 

 

What i want to do, for given p is set an upper limit N for k then check to see if R_kpR_k is prime for values of k from 1 to N

The bit I can’t handle is expressing the “wrapped” number for a given p and k, and also indexing from k to k+1, until N is reached or a prime found.

Any assistance gratefully received.

Best regards

David.

 

 

 

 

 

Hi, 

I am interested to find the primes in in field Zp which Have prime inverses.  In particular, for any prime p up to an arbitrary number N, the number of pairs (a,b) where a and b are mutually inverse primes <p in Zp. Obviously in this context (a,b) is the same as (b,a) so no need for double counting. What I have so far is the following simple code for finding the inverse of prime q<p, given a chosen value of p. I can then see which results are prime and which are composite.

N:=p:

for n from 1 to N do

X:=q*n-1;

if mod(X,p)=0 then print(n);

end if:

end do:

This “hand cranking” method works but is of course boring but I don’t know how to scale it up to a more efficient code as described above. I would be grateful to anyone  able to assist me with this.

Thanks in advance

David

ithprime(n)  gives the unique prime whose index is n, thus ithprime(10) returns 29. Question: is there an inverse function in Maple, which given prime p would return n such that ithprime(n) = p?

Or if no such function exists can anyone suggest a way such a function could be coded?

Thanks in advance

David.

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