David Sycamore

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1 years, 190 days

MaplePrimes Activity

These are questions asked by David Sycamore

How is it possible in Maple to keep hold of pre determined results for comparison with subsequent results so that a recursive decision can be made to either modify the list of “kept” data or to continue to the next calculation, etc... ?

Example: a simple “subtract or double” sequence. If subtracting (say 1) from the current number would result in a term we already have,  then double it instead, and start over again with subtraction.

Formally: a(0)=0, a(1)=1, and for n>=1,

a(n+1) = a(n)-1 if that number has not been found already, else a(n+1)=2*a(n).


The arithmetic operations are facile but how to organise the keeping and comparison process??



Take any odd number k, with distinct odd prime divisors p_1,...p_k and ask the following question: What is the smallest even number which when added to each prime divisor p_i of k, gives another prime? Example: k= 119=7*17, so the smallest even number is 6 because 7+6=11 and 17+6=23. NB: some numbers (eg 105,195,231...) seem to have no solution. I would appreciate assistance with a code to calculate for every odd k, the even number in question, or to allocate 0 for numbers where no solution has been found up to some suitably convincing high number like N= 10^5 or 10^6, (a parameter I could change if desired). The idea would be to conjecture that the apparently no solution numbers (up to N) don’t actually have a solution. Any assistance much appreciated in advance. I have found the solutions up to k=781 by hand, but my hand is getting tired now.



ps: I would like to have the option (if possible) to output the numbers with solution =  0, and the numbers which are the smallest to have solution 2*n, as separate sequences, for n up to some arbitrary value (ie 3,7,23,69,93...). (n>=1). Hope this is not asking too much. 

n is a Carmichael number iff for every prime factor p of n, p-1/n-1.

Question: How to find odd squarefree composite numbers n having k distinct prime divisors, and the property that exactly k-1 prime divisors satisfy the Carmichael requirement, p-1/n-1 ?

Examples: 231,1045,1635. In these cases k=3 and the prime divisors satisfying the criteria are the greatest and smallest. I have a code for this but would like to compute the general case, where the criteria is satisfied for precisely any k-1 divisors.

Any assistance greatly appreciated.



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.


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



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:


for k from 1 to N do


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





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