bash, 43 bytes

factor $1|awk '{print $2-$3&&$3-$4&&NF==4}' 

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Input via command line argument, outputs 0 or 1 to stdout.

Fairly self-explanatory; parses the output of factor to check that the first and second factors are different, the second and third are different (they're in sorted order, so this is sufficient), and there are four fields (the input number and the three factors).

MATL, 7 bytes

_YF7BX= 

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Explanation

_YF % Implicit input. Nonzero exponents of prime-factor decomposition 7 % Push 7 B % Convert to binary: gives [1 1 1] X= % Is equal? Implicit display 

C, 88 78 126 58 77 73 + 4 (lm) = 77 bytes

l,j;a(i){for(l=1,j=0;l++<i;fmod(1.*i/l,l)?i%l?:(i/=l,j++):(j=9));l=i==1&&j==3;} 

Ungolfed commented explanation:

look, div; //K&R style variable declaration. Useful. Mmm. a ( num ) { // K&R style function and argument definitions. for ( look = 1, div = 0; // initiate the loop variables. look++ < num;) // do this for every number less than the argument: if (fmod(1.0 * num / look, look)) // if num/look can't be divided by look: if( !(num % look) ) // if num can divide look num /= look, div++; // divide num by look, increment dividers else div = 9; // if num/look can still divide look // then the number's dividers aren't unique. // increment dividers number by a lot to return false. // l=j==3; // if the function has no return statement, most CPUs return the value // in the register that holds the last assignment. This is equivalent to this: return (div == 3); // this function return true if the unique divider count is 3 } 

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CJam, 11 bytes

rimFz1=7Yb= 

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Explanation

Based on my MATL answer.

ri e# Read integer mF e# Factorization with exponents. Gives a list of [factor exponent] lists z e# Zip into a list of factors and a list of exponents 1= e# Get second element: list of exponents 7 e# Push 7 Yb e# Convert to binary: gives list [1 1 1] = e# Are the two lists equal? Implicitly display 

Jelly, 8 bytes

ÆEḟ0⁼7B¤ 

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Uses Luis Mendo's algorithm.

Explanation:

ÆEḟ0⁼7B¤ ÆE Prime factor exponents ḟ0 Remove every 0 ⁼7B¤ Equal to 7 in base 2? 

Jelly, 6 bytes

ÆE²S=3 

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How it works

ÆE²S=3 Main link. Argument: n ÆE Compute the exponents of n's prime factorization. ² Take their squares. S Take the sum. =3 Test the result for equality with 3. 

Husk, 6 bytes

≡ḋ3Ẋ≠p 

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Returns 1 for sphenic numbers and 0 otherwise.

Explanation

≡ḋ3Ẋ≠p Example input: 30 p Prime factors: [2,3,5] Ẋ≠ List of absolute differences: [1,2] ≡ Is it congruent to... ? ḋ3 the binary digits of 3: [1,1] 

In the last passage, congruence between two lists means having the same length and the same distribution of truthy/falsy values. In this case we are checking that our result is composed by two truthy (i.e. non-zero) values.

Ruby, 81 49 46 bytes

Includes 6 bytes for command line options -rprime.

->n{n.prime_division.map(&:last)==[1]*3} 

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Mathematica, 31 bytes

SquareFreeQ@#&&PrimeOmega@#==3& 

05AB1E, 7 5 bytes

ÓnO3Q 

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Uses Dennis's algorithm.

Haskell, 59 bytes

f x=7==sum[6*0^(mod(div x a)a+mod x a)+0^mod x a|a<-[2..x]] 

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Pyth, 9 bytes

&{IPQq3lP 

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Actually, 7 bytes

w♂N13α= 

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Explanation:

w♂N13α= w Push [prime, exponent] factor pairs ♂N Map "take last element" 1 Push 1 3 Push 3 α Repeat = Equal? 

Japt, 14 bytes

k k@è¥X ÉÃl ¥3 

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J, 15 bytes

7&(=2#.~:@q:)~* 

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Explanation

7&(=2#.~:@q:)~* Input: integer n * Sign(n) 7&( )~ Execute this Sign(n) times on n If Sign(n) = 0, this returns 0 q: Prime factors of n ~:@ Nub sieve of prime factors 2#. Convert from base 2 = Test if equal to 7 

Dyalog APL, 26 bytes

⎕CY'dfns' ((≢,∪)≡3,⊢)3pco⎕ 

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Mathematica, 44 bytes

Plus@@Last/@#==Length@#==3&@FactorInteger@#& 

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Python 3, 54 53 bytes

lambda n:sum(1>>n%k|7>>k*k%n*3for k in range(2,n))==6 

Thanks to @xnor for golfing off 1 byte!

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C, 91 102 bytes, corrected (again), golfed, and tested for real this time:

<strike>s(c){p,f,d;for(p=2,f=d=0;p<c&&!d;){if(c%p==0){c/=p;++f;if(c%p==0)d=1;}++p;}c==p&&f==2&&!d;}</strike> s(c){int p,f,d;for(p=2,f=d=0;p<c&&!d;){if(c%p==0){c/=p;++f;if(c%p==0)d=1;}++p;}return c==p&&f==2&&!d;} 

/* This also works in 93 bytes, but since I forgot about the standard rules barring default int type on dynamic variables, and about the not allowing implicit return values without assignments, I'm not going to take it:

p,f,d;s(c){for(p=2,f=d=0;p<c&&!d;){if(c%p==0){c/=p;++f;if(c%p==0)d=1;}++p;}p=c==p&&f==2&&!d;} 

(Who said I knew anything about C? ;-)

Here's the test frame with shell script in comments:

/* betseg's program for sphenic numbers from */ #include <stdio.h> #include <stdlib.h> #include <limits.h> #include <math.h> /* compile with -lm */ /* l,j;a(i){for(l=1,j=0;l<i;i%++l?:(i/=l,j++));l=i==1&&j==3;} */ #if defined GOLFED l,j;a(i){for(l=1,j=0;l++<i;fmod((float)i/l,l)?i%l?:(i/=l,j++):(j=9));l=i==1&&j==3;} #else int looker, jcount; int a( intval ) { for( looker = 1, jcount = 0; looker++ < intval; /* Watch odd intvals and even lookers, as well. */ fmod( (float)intval/looker, looker ) ? intval % looker /* remainder? */ ? 0 /* dummy value */ : ( inval /= looker, jcount++ /* reduce the parameter, count factors */ ) : ( jcount = 9 /* kill the count */ ) ) /* empty loop */; looker = intval == 1 && jcount == 3; /* reusue looker for implicit return value */ } #endif /* for (( i=0; $i < 100; i = $i + 1 )) ; do echo -n at $i; ./sphenic $i ; done */ 

I borrowed betseg's previous answer to get to my version.

This is my version of betseg's algorithm, which I golfed to get to my solution:

/* betseg's repaired program for sphenic numbers */ #include <stdio.h> #include <stdlib.h> #include <limits.h> int sphenic( int candidate ) { int probe, found, dups; for( probe = 2, found = dups = 0; probe < candidate && !dups; /* empty update */ ) { int remainder = candidate % probe; if ( remainder == 0 ) { candidate /= probe; ++found; if ( ( candidate % probe ) == 0 ) dups = 1; } ++probe; } return ( candidate == probe ) && ( found == 2 ) && !dups; } int main( int argc, char * argv[] ) { /* Make it command-line callable: */ int parameter; if ( ( argc > 1 ) && ( ( parameter = (int) strtoul( argv[ 1 ], NULL, 0 ) ) < ULONG_MAX ) ) { puts( sphenic( parameter ) ? "true" : "false" ); } return EXIT_SUCCESS; } /* for (( i=0; $i < 100; i = $i + 1 )) ; do echo -n at $i; ./sphenic $i ; done */ 

Javascript (ES6), 87 bytes

n=>(a=(p=i=>i>n?[]:n%i?p(i+1):[i,...p(i,n/=i)])(2)).length==3&&a.every((n,i)=>n^a[i+1]) 

Example code snippet:

f= n=>(a=(p=i=>i>n?[]:n%i?p(i+1):[i,...p(i,n/=i)])(2)).length==3&&a.every((n,i)=>n^a[i+1]) for(k=0;k<10;k++){ v=[30,121,231,154,4,402,79,0,60,64][k] console.log(`f(${v}) = ${f(v)}`) }

Python 2, 135 121 bytes

• Quite long since this includes all the procedures: prime-check, obtain-prime factors and check sphere number condition.

lambda x:(lambda t:len(t)>2and t[0]*t[1]*t[2]==x)([i for i in range(2,x)if x%i<1and i>1and all(i%j for j in range(2,i))]) 

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Python 2, 59 bytes

lambda x:6==sum(5*(x/a%a+x%a<1)+(x%a<1)for a in range(2,x)) 

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J, 23 bytes

0:`((~.-:]*.3=#)@q:)@.* 

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Handling 8 and 0 basically ruined this one...

q: gives you all the prime factors, but doesn't handle 0. the rest of it just says "the unique factors should equal the factors" and "the number of them should be 3"

PHP, 66 bytes:

for($p=($n=$a=$argn)**3;--$n;)$a%$n?:$p/=$n+!++$c;echo$c==7&$p==1; 

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Infinite loop for 0; insert $n&& before --$n to fix.

breakdown

for($p=($n=$a=$argn)**3; # $p = argument**3 --$n;) # loop $n from argument-1 $a%$n?: # if $n divides argument $p/=$n # then divide $p by $n +!++$c; # and increment divisor count echo$c==7&$p==1; # if divisor count is 7 and $p is 1, argument is sphenic 

example
argument = 30:
prime factors are 2, 3 and 5
other divisors are 1, 2*3=6, 2*5=10 and 3*5=15
their product: 1*2*3*5*6*10*15 is 27000 == 30**3

VB.NET (.NET 4.5), 104 bytes

Function A(n) For i=2To n If n Mod i=0Then A+=1 n\=i End If If n Mod i=0Then A=4 Next A=A=3 End Function 

I'm using the feature of VB where the function name is also a variable. At the end of execution, since there is no return statement, it will instead pass the value of the 'function'.

The last A=A=3 can be thought of return (A == 3) in C-based languages.

Starts at 2, and pulls primes off iteratively. Since I'm starting with the smallest primes, it can't be divided by a composite number.

Will try a second time to divide by the same prime. If it is (such as how 60 is divided twice by 2), it will set the count of primes to 4 (above the max allowed for a sphenic number).

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JavaScript (ES6), 80 bytes

n=>(p=(n,f=2)=>n%f?p(n,f+1):f,(a=p(n))<n&&(b=p(n/=a))<n&&(c=p(n/=b))==n&a<b&b<c) 

Using a recursive function to get the smaller factor.
Output 1 if sphenic, false if there are 2 or less factors and 0 otherwise

Test

F= n=>(p=(n,f=2)=>n%f?p(n,f+1):f,(a=p(n))<n&&(b=p(n/=a))<n&&(c=p(n/=b))==n&a<b&b<c) ;[30,121,231,154,4,402,79,0,60,64,8,210].forEach( x=>console.log(x,F(x)) )

Dyalog APL, 51 49 48 46 45 43 bytes

1∊((w=×/)∧⊢≡∪)¨(⊢∘.,∘.,⍨){⍵/⍨2=≢∪⍵∨⍳⍵}¨⍳w←⎕ 

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I wanted to submit one that doesn't rely on the dfns namespace whatsoever, even if it is long.

J, 15 14 19 bytes

Previous attempt: 3&(=#@~.@q:)~*

Current version: (*/*3=#)@~:@q: ::0:

How it works:

(*/*3=#)@~:@q: ::0: Input: integer n ::0: n=0 creates domain error in q:, error catch returns 0 q: Prime factors of n ~:@ Nub sieve of prime factors 1 for first occurrence 0 for second (*/*3=#)@ Number of prime factors is equal to 3, times the product across the nub sieve (product is 0 if there is a repeated factor or number of factors is not 3) 

This passes for cases 0, 8 and 60 which the previous version didn't.

Regex (ECMAScript), 46 bytes

^((?=(xx+?)\2*$)(?=(x+)(\3+$))\4(?!\2+$)){3}x$ 

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This works similarly to Match strings whose length is a fourth power and very similarly to part of Is this a consecutive-prime/constant-exponent number:

^ ( # Loop the following: (?=(xx+?)\2*$) # \2 = smallest prime factor of tail (?= (x+)(\3+$) # \3 = tail / {smallest prime factor of tail}; \4 = tool to make tail = \3 )\4 # tail = \3 (?!\2+$) # assert that tail is no longer divisible by \2 ){3} # Execute the loop exactly 3 times x$ # assert tail == 1