This question already has an answer here:
The primorial of a number is the product of all the primes until that number, itself included.
Take a number from STDIN as the input and evaluate its primorial, also known as the prime factorial.
Don't use libraries. Shortest code wins.
Test cases:
1 1
2 2
3 6
4 6
5 30
6 30
7 210
8 210
9 210
10 210
11 2310
12 2310
13 30030
14 30030
15 30030
16 30030
17 510510
18 510510
19 9699690
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*/p:i._1 p:1+
Examples:
*/p:i._1 p:1+7
210
*/p:i._1 p:1+12
2310
P=k=1
exec"P*=P**k%k<1or k;k+=1;"*input()
print P
Counts up k from 1 to the input, updating the primorial P by multiplying P by k whenever it's prime. Since P contains all lower prime factors, a sufficiently high power of P will contain k as a factor if and only if k is non-prime. The power P**k suffices, and we check whether P**k%k<1, multipling P by 1 if so (doing nothing) and by k otherwise.
Times@@Prime@Range@PrimePi@
Usage
Times@@Prime@Range@PrimePi@100
(*
2305567963945518424753102147331756070
*)
@ symbol at the end meant to be there? On it's own that'd be invalid syntax. You could probably just omit it.
\$\endgroup\$
– numbermaniac
Oct 2 '17 at 0:46
The large Pi represents the product function.
PrimePi@n returns the number of primes up to and including n.
Prime@i returns the ith prime.
Quickly golfed, I imagine there are better python solutions.
n=t=1;exec"t*=(1,n)[all(n%i for i in range(2,n))];n+=1;"*input();print t
I used wikipedia to verify up to 71
Usage:
$ python primorial.py
100
2305567963945518424753102147331756070
$ python primorial.py
101
232862364358497360900063316880507363070
prodeuler(x=1,input,x)
main=getLine>>= \v->print.product.takeWhile(<=read v)$[n|n<-[2..],all((>0).rem n)[2..n-1]]
or (also 90 chars):
main=getLine>>=print.product.flip takeWhile [n|n<-[2..],all((>0).rem n)[2..n-1]].(>=).read
ghc primorial.hs
$ echo "12" | ./primorial
2310
or
$ ./primorial
12<ENTER>
2310
takeWhile is unnecessary, we can just add an upper limit for the [2..] instead. Instead of getLine + read, we can use readLn. Also, do-notation is slightly shorter than using >>=. Result: main=do v<-readLn;print$product[n|n<-[2..v],all((>0).rem n)[2..n-1]] (68 characters)
\$\endgroup\$
– hammar
Jun 18 '13 at 21:10
x=1:scan();prod(x[rowSums(outer(x,x,`%%`)<1)<3])
p(n,x){return x==1?n:n%x?p(n,x-1):1;}
f(n){return n==1?1:p(n,n-1)*f(n-1);}
main(){int n;scanf("%d",&n);printf("%d\n",f(n));}
Smalltalk, Squeak 4.x flavour
Let's say we read n from stdin in 38 chars (this kind of requirement is just killing!)
n:=FileStream stdin nextLine asNumber.
If only product were defined in Collection in base image, this could be as simple as
^(0class primesUpTo:n+1)product
or:
^((2to:n)select:#isPrime)product
Unfortunately, we have to code #product by ourself in 48 chars (86 total).
(0class primesUpTo:n+1)inject:1into:[:p :i|p*i]
Or if primesUpTo: is considered as a library, we can use isPrime in 62 chars (100 total)
^(1to:n)inject:1into:[:p :i|i isPrime ifTrue:[p*i]ifFalse:[p]]
If even isPrime is forbidden we can find by ourselves in 67 chars (105 total)
p:=1. 2to:n do:[:i|((2to:i-1)anySatisfy:[:d|i\\d=0])or:[p:=p*i]].^p
Or for the fun, we can use a sieve in 108 chars (146 total)
s:=(p:=y:=1)<<n-2. 2to:n do:[:x|y:=(z:=y)*2+1.(z+1bitAnd:s)=0or:[p:=p*x.s:=z<<(n//x*x+x)//y+z+1bitAnd:s]].^p
Explanation of the sieve at http://smallissimo.blogspot.fr/2011/07/revisiting-sieve-of-erathostenes.html
Nota: bitAnd: has a short-cut operator & in Pharo 2.0, which would reduce Sieve cost to 94 chars (132 total)
s:=(p:=y:=1)<<n-2. 2to:n do:[:x|y:=(z:=y)*2+1.z+1&s=0or:[p:=p*x.s:=z<<(n//x*x+x)//y+z+1&s]].^p
main=interact$ \n->show$product[x|x<-[2..read n],and[x`mod`y>0|y<-[2..x-1]]]
Example:
$ echo 12 | ./primorial
2310
class M{public static void main(String[]a){c(new Long(a[0]));}static void c(long n){long r=1,i=0;for(;++i<=n;r*=p(i)?i:1);System.out.print(r);}static boolean p(long n){int i=2;while(i<n)n=n%i++<1?0:n;return n>1;}}
Ungolfed:
class M{
static void c(long n){
long r = 1,
i = 0;
for (; ++i <= n; r *= p(i) ? i : 1);
System.out.print(r);
}
public static void main(String[] a){
c(new Long(a[0]));
}
static boolean p(long n){
int i = 2;
while(i < n){
n = n % i++ < 1 ? 0 : n;
}
return n > 1;
}
}
Usage: java -jar M.jar 17
Output: 510510
S²P¬-*
S²P¬-*
S # Produce as stack from 1 to the input.
² - # Filter it by the function provided
P¬ # Is it not a prime?
* # Get the product of the now list of primes.
prod(seq(piecewise(isPrime(x),x,1),x,1,n
Since the calculator doesn't have a "select" function of any kind, a piecewise/hybrid function is used to return the number if it's prime, otherwise return 1. seq does this over a range of integers from 1 to n, and prod multiplies the whole list together.
40 bytes for the function, +1 byte to enter n in the parameters box.
