Function to find the next prime number...

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Jyothi Alugolu
Jyothi Alugolu on 19 Feb 2018
Commented: Stephen on 28 Jul 2021
I have a vector of size 1*152.Now i want to find the next prime number of every number present in the vector..
Ex: My vector is a=[2 4 7 8] i want the output as [2 5 7 11]..i.e., if the number is a prime then that number will be the output i.e., like 2 and 7 in the given example...
I tried using nextprime like below it gives the following error:
case 1:
>>nextprime(sym(100))
Undefined function 'nextprime' for input arguments of type 'sym'.
case 2:
>> nextprime(3)
Undefined function 'nextprime' for input arguments of type 'double'.

Accepted Answer

Basil C.
Basil C. on 19 Feb 2018
Method 1 This functionality does not run in MATLAB and can be used only via MuPAD Notebook Interface.
  • To create an MuPAD interface use the following code
mupad
nb = allMuPADNotebooks
Then a interface screen shall pop up where you can proceed by using the nextprime(num) function.
Method 2
  • You could also create a user defined function to compute the next prime number. This function takes only a non-negative integers as an argument
function p = nextprime(n)
if (isprime(n))
p=n;
else
while(~isprime(n))
n=n+1;
end
p=n;
end
end
  4 Comments
Hicham Satti
Hicham Satti on 31 Aug 2020
%That will run very well
function k=next_prime(n)
if ~n>0 || n~=fix(n) || ~isscalar(n)
error ('Tap a integer positif scalar argument');
else
if isprime(n+1)
k=n+1;
else
j=n+1;
while ~isprime(j)
k=j+1;
j=j+1;
end
end
end

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More Answers (12)

Arafat Roney
Arafat Roney on 11 May 2020
function k=next_prime(n)
i=n+1;
if(isprime(i))
k=i;
else
while(~isprime(i))
i=i+1;
end
k=i;
end
end

Walter Roberson
Walter Roberson on 19 Feb 2018
nextprime() was added to the Symbolic Toolbox in R2016b.
In releases before that,
feval(symengine, 'nextprime', sym(100))

Siddharth Joshi
Siddharth Joshi on 25 Apr 2020
function k = next_prime(n)
if (~isscalar(n) || n<1 || n ~= fix(n))
error('n should be positive scalar ineger!!!')
else
p=-1;
while p<=0
n=n+1;
p=isprime(n)
end
k=n
end
end
k = next_prime(79)
k =
83
  4 Comments
Stephen
Stephen on 28 Jul 2021
"The code would have been better as"
... and ultimately simplifies down to this.

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Buwaneka Dissanayake
Buwaneka Dissanayake on 21 Jun 2020
% what's wrong with this? it take too long to run & fail
function n = next_prime(n)
k = n+1;
while ~isprime(k)
n = n+1;
end
end
what's wrong with this? it take too long to run & fail.
  2 Comments
SAKSHI CHANDRA
SAKSHI CHANDRA on 22 Jul 2020
your control statement defines k but there is nothing related in the block statements which would check
~isprime(k)

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MD SADIQUE IQBAL
MD SADIQUE IQBAL on 17 Jul 2020
unction n = next_prime(n)
k = n+1;
while ~isprime(k)
n = n+1;
end
end
  2 Comments
SAKSHI CHANDRA
SAKSHI CHANDRA on 22 Jul 2020
your control statement defines k but there is nothing related in the block statements which would check
~isprime(k)

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SAKSHI CHANDRA
SAKSHI CHANDRA on 22 Jul 2020
function k = nxt_prime(n)
k=n+1;
while ~isprime(k)
k=k+1;
end
end

Ravindra Pawar
Ravindra Pawar on 13 Aug 2020
Edited: Ravindra Pawar on 13 Aug 2020
function k = next_prime(n) %function definition
while ~isprime(n+1) %if n+1 is prime we are out of for loop else loop restarts
n = n+1;
end
k = n+1;
end

shweta s
shweta s on 14 Aug 2020
%to find the next prime no.
function p = next_prime(n)
if (isprime(n))
p=n+1;
else
while(~isprime(n))
n=n+1;
end
p=n;
end
end

Hicham Satti
Hicham Satti on 31 Aug 2020
%That will run very well
function k=next_prime(n)
if ~n>0 || n~=fix(n) || ~isscalar(n)
error ('Tap a integer positif scalar argument');
else
if isprime(n+1)
k=n+1;
else
j=n+1;
while ~isprime(j)
k=j+1;
j=j+1;
end
end
end
  3 Comments
Rik
Rik on 8 Sep 2020
Because I'm just one person trying to clean up thread like this. And you didn't answer my question (neither here, nor on the other next_prime thread).

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Pragyan Dash
Pragyan Dash on 19 Sep 2020
function k = next_prime(n)
while (~isprime(n + 1))
n = n + 1;
end
k = n + 1;
end

Malgorzata Frydrych
Malgorzata Frydrych on 26 Jun 2021
function k= next_prime(n)
if ~isscalar(n) || n<=0 || mod(n,1)~=0;
error('number should be a positive integer scalar')
end
k=0;
while ~isprime(k)
n=n+1;
k=n;
end
end
  1 Comment
Walter Roberson
Walter Roberson on 26 Jun 2021
Is this efficient? If you are currently at 15, is there a point in testing 16?

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Dikshita Madkatte
Dikshita Madkatte on 14 Jul 2021
Edited: Dikshita Madkatte on 14 Jul 2021
function k=next_prime(n)
if ~n>0 || n~=fix(n) || ~isscalar(n);
fprintf('n should be positive interger')
end
i=n+1;
if (isprime(i))
k=i;
else
while(~isprime(i))
i=i+1;
end
k=i;
end
end
  1 Comment
Rik
Rik on 14 Jul 2021
A few remarks:
fprintf is not an error. Your code will still run after it fails the check.
You can increment i by two, since 2 is the only even prime, and the while loop will not be reached if n is 1.
You forgot to write documentation for your function. What is this going to teach? Why should it not be deleted?

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