Pretty Smooth Moves

Question

In arithmetic, an n-smooth number, where n is a given prime number, is mathematically defined as a positive integer that has no prime factors greater than n. For example, 42 is 7-smooth because all its prime factors are less than or equal to 7, but 44 is not 7-smooth because it also has 11 as a prime factor.

Define a pretty smooth number as a number with no prime factors greater than its own square root. Thus, the list of pretty smooth numbers can be formulated as follows:

• (EDITED!) 1 is a pretty smooth number, due to its complete lack of any prime factors. (Note that in the original version of this question, 1 was erroneously excluded from the list, so if you exclude it from your outputs you won't be marked wrong.)
• Between 4 (= 2²) and 8, the pretty smooth numbers are 2-smooth, meaning they have 2 as their only prime factor.
• Between 9 (= 3²) and 24, the pretty smooth numbers are 3-smooth, and can have 2s and 3s in their prime factorizations.
• Between 25 (= 5²) and 48, the pretty smooth numbers are 5-smooth, and can have 2s, 3s, and 5s in their prime factorizations.
• And so on, upgrading the criteria every time the square of the next prime number is reached.

The list of pretty smooth numbers is fixed, and begins as follows: 1, 4, 8, 9, 12, 16, 18, 24, 25, ...

Your challenge is to write code that will output all pretty smooth numbers up to and including 10,000 (= 100²). There must be at least one separator (it doesn't matter what kind -- space, comma, newline, anything) between each number in the list and the next, but it is completely irrelevant what character is used.

As per usual, lowest byte count wins -- obviously, simply outputting the list isn't going to be too beneficial to you here. Good luck!

Answer 1

Jelly, 12 bytes

Æf>½S
³²ḊÇÐḟ

Try it online!

How it works

³²ḊÇÐḟ  Main link. No arguments.

³       Yield 100.
 ²      Square it to yield 10,000.
  Ḋ     Dequeue; yield [2, ..., 10,000].
   ÇÐḟ  Filter-false; keep elements for which the helper link returns 0.

Æf>½S   Helper link. Argument: n

Æf      Compute the prime factorization of n.
  >½    Compare the prime factors with the square root of n.
    S   Sum; add the resulting Booleans.

Answer 2

Brachylog, 21 19 bytes

1 byte thanks to Fatalize, for inspiration of another 1 byte.

100^:4reP$ph^<=P@w\

Try it online!

Takes about 6 seconds here.

100^:4reP$ph^<=P@w\
100                      100
   ^                     squared
    :4                   [10000,4]
      r                  [4,10000]
       eP                P is an integer in that interval (choice point),
        P$ph^<=P         P, prime factorized (from biggest to smallest),
                         take the first element, squared, is less than
                         or equal to P
               P@w       Write P with a newline,
                  \      Backtrack to the last choice point and make
                         a different choice until there is no more
                         choice and the program halts.

Previous 21-byte solution

100^:4reP'($pe^>P)@w\

Try it online!

Takes about 6 seconds here.

100^:4reP'($pe^>P)@w\
100                      100
   ^                     squared
    :4                   [10000,4]
      r                  [4,10000]
       eP                P is an integer in that interval (choice point),
        P'(      )       The following about P cannot be proved:
           $pe               one of its prime factor
              ^              squared
               >P            is greater than P
                  @w     Write P with a newline,
                    \    Backtrack to the last choice point and make
                         a different choice until there is no more
                         choice and the program halts.

Answer 3

Haskell, 53 bytes

r=[1..10^4]
[n|n<-r,product[x|x<-r,x*x<=n]^n`mod`n<1]

I don't have time to golf this now, but I want to illustrate a method for testing if n is pretty smooth: Multiply the numbers from 1 to sqrt(n) (i.e. compute a factorial), raise the product to a high power, and check if the result is a multiple of n.

Change to r=[2..10^4] if 1 should not be output.

Answer 4

Pyth, 16 15 bytes

1 byte thanks to Jakube.

tf!f>*YYTPTS^T4

Try it online!

tf!f>*YYTPTS^T4
             T   10
            ^T4  10000
           S^T4  [1,2,3,...,10000]
 f               filter for elements as T for
                 which the following is truthy: 
         PT          prime factorization of T
   f                 filter for factor as Y:
     *YY                 Y*Y
    >   T                greater than T ?
  !                  logical negation
t                remove the first one (1)

Answer 5

05AB1E, 16 14 13 bytes

4°L¦vyf¤yt›_—

Explanation

4°L¦v             # for each y in range 2..10000
      yf¤         # largest prime factor of y
         yt       # square root of y
           ›_     # less than or equal
             —    # if true then print y with newline

Try it online

Answer 6

Matlab, 58 57 56 52 48 bytes

for k=1:1e4
if factor(k).^2<=k
disp‌​(k)
end
end

For each number it checks if all factors squared are not larger than the number itself. If yes, displays that number.

Thanks to @Luis Mendo for golfing this approach

---

Another approach (50 bytes):

n=1:10^4;for k=n
z(k)=max(factor(k))^2>k;end
n(~z)

For each number computes whether its maximum prime factor squared is less than the number itself. Then uses it for indexing.

Answer 7

Actually, 11 bytes

4╤R`;yM²≤`░

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Does not include 1.

Explanation:

4╤R`;yM²≤`░
4╤R          range(10**4)
   `;yM²≤`░  filter: take values where
    ;yM²       the square of the largest prime factor
        ≤      is less than or equal to the value

Answer 8

SQF, 252 227 220

Standard script format:

#define Q(A,B) for #A from 2 to B do{
Q(i,10000)if([i]call{params["j"];u=sqrt j;a=true;Q(k,u)a=a and((j%k!=0)or(j/k<u)or!([j/k]call{params["x"];q=true;Q(z,sqrt x)q=q and(x%z!=0)};q}))};a})then{systemChat format["%1",i]}}

Include the pre-processor in the compilation chain when calling eg:

• execVM "FILENAME.sqf"
• call compile preprocessFile "FILENAME.sqf"

This writes to the System Chat log, which is the closest thing SQF has to stdout

Answer 9

C, 113 bytes

#include<stdio.h>
main(a){for(;++a<10001;){int n=2,c=a;for(;n*n<=a;n++)while(c%n<1)c/=n;if(c<2)printf("%d ",a);}}

Ideone it!

Answer 10

Pyke, 13 12 11 bytes

T4^S#DP#X<!

Try it here!

(Link only goes up to 10^3 because 10^4 times out)

T4^S        -  one_range(10^4)
    #DP#X<! - filter_true(V, ^): (as i)
      P     -   factors(i)
       #X<! -  filter_true(V, ^):
        X   -   ^ ** 2
         <! -    not (i < ^)

Answer 11

J, 20 bytes

(#~{:@q:<:%:)2+i.1e4

Result:

   (#~{:@q:<:%:)2+i.1e4
4 8 9 12 16 18 24 25 27 30 32 36 40 45 48 49 50 54 56 60 63 64 70 72 75 80...

Try it online here.

Answer 12

Python 2, 90 bytes

for i in range(4,10001):
 n=2;j=i
 while n*n<=j:
  while i%n<1:i/=n
  n+=1
 if i<2:print j

Ideone it!

Answer 13

R, 97 bytes

b=F;for(j in 1:1e4){for(i in which(!j%%1:j)[-1])if(which(!i%%1:i)[2]==i)b=i<=j^0.5;if(b)print(j)}

ungolfed

b <- F                               #Initializes
for (j in 1:1e4){                    #Loop across integers 1..10^4
    a <- which(!j%%1:j)[-1]          #Finds all factors
    for (i in a)                     #Loop across factors
         b <- which(!i%%1:i)[2]==i   #Tests primeness
         if(b) c <- i<=j^0.5         #If prime, tests smoothness
    if(c) print(j)                   #If biggest prime factor gives smooth
}                                    #result, Prints the number.

Answer 14

Pyth, 12 bytes

g#^ePT2tS^T4

Does not include 1.