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Haskell, 59 bytes

n%p|n*n<2=0|mod n p>0=n%(p+1)|r<-div n p=r+p*r%2
(%2)

Implements the recursive definition directly, with an auxiliary variable p that counts up to search for potential prime factors, starting from 2. The last line is the main function, which plugs p=2 to the binary function defined in the first line.

The function checks each case in turn:

  • If n*n<2, then n is one of -1,0,1, and the result is 0.
  • If n is not a multiple of p, then increment p and continue.
  • Otherwise, express n=p*r, and by the "derivative" property, the result is r*a(p)+p*a(r), which simplifies to r+p*a(r) because p is prime.

The last case saves bytes by binding r in a guardbinding r in a guard, which also avoids the 1>0 for the boilerplate otherwise. If r could be bound earlier, the second condition mod n p>0 could be checked as r*p==n, which is 3 bytes shorter, but I don't see how to do that.

Haskell, 59 bytes

n%p|n*n<2=0|mod n p>0=n%(p+1)|r<-div n p=r+p*r%2
(%2)

Implements the recursive definition directly, with an auxiliary variable p that counts up to search for potential prime factors, starting from 2. The last line is the main function, which plugs p=2 to the binary function defined in the first line.

The function checks each case in turn:

  • If n*n<2, then n is one of -1,0,1, and the result is 0.
  • If n is not a multiple of p, then increment p and continue.
  • Otherwise, express n=p*r, and by the "derivative" property, the result is r*a(p)+p*a(r), which simplifies to r+p*a(r) because p is prime.

The last case saves bytes by binding r in a guard, which also avoids the 1>0 for the boilerplate otherwise. If r could be bound earlier, the second condition mod n p>0 could be checked as r*p==n, which is 3 bytes shorter, but I don't see how to do that.

Haskell, 59 bytes

n%p|n*n<2=0|mod n p>0=n%(p+1)|r<-div n p=r+p*r%2
(%2)

Implements the recursive definition directly, with an auxiliary variable p that counts up to search for potential prime factors, starting from 2. The last line is the main function, which plugs p=2 to the binary function defined in the first line.

The function checks each case in turn:

  • If n*n<2, then n is one of -1,0,1, and the result is 0.
  • If n is not a multiple of p, then increment p and continue.
  • Otherwise, express n=p*r, and by the "derivative" property, the result is r*a(p)+p*a(r), which simplifies to r+p*a(r) because p is prime.

The last case saves bytes by binding r in a guard, which also avoids the 1>0 for the boilerplate otherwise. If r could be bound earlier, the second condition mod n p>0 could be checked as r*p==n, which is 3 bytes shorter, but I don't see how to do that.

deleted 3 characters in body
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xnor
xnor
  • 146.6k
  • 26
  • 279
  • 652

Haskell, 59 bytes

n%p|n*n<2=0|mod n p>0=n%(p+1)|r<-div n p=r+p*r%2
(%2)

The last line is the main function, with the first line defining an auxiliary function. Implements the recursive definition directly, with an auxiliary variable p that counts up to search for potential prime factors, starting from 2. The last line is the main function, which plugs p=2 to the binary function defined in the first line.

The function checks each case in turn:

  • If n*n<2, then n is one of -1,0,1, and the result is 0.
  • If n is not a multiple of p, then increment p and recursecontinue.
  • Otherwise, express n=p*r, and by the "derivative" property, the result is r*a(p)+p*a(r), which simplifies to r+p*a(r) because p is prime.

For theThe last case, we save saves bytes by binding r in a guard, while atwhich also avoids the same time saving on 1>0 for the boilerplate otherwise. If r could be bound earlier, the second condition mod n p>0 could be checked as r*p==n, which is 3 bytes shorter, but I don't knowsee how to do that.

Haskell, 59 bytes

n%p|n*n<2=0|mod n p>0=n%(p+1)|r<-div n p=r+p*r%2
(%2)

The last line is the main function, with the first line defining an auxiliary function. Implements the recursive definition directly, with an auxiliary variable p that counts up to search for potential prime factors, starting from 2.

The function checks each case in turn:

  • If n*n<2, then n is one of -1,0,1, and the result is 0.
  • If n is not a multiple of p, then increment p and recurse.
  • Otherwise, express n=p*r, and by the "derivative" property, the result is r*a(p)+p*a(r), which simplifies to r+p*a(r) because p is prime.

For the last case, we save bytes by binding r in a guard, while at the same time saving on 1>0 for the boilerplate otherwise. If r could be bound earlier, the second condition mod n p>0 could be checked as r*p==n, which is 3 bytes shorter, but I don't know how to do that.

Haskell, 59 bytes

n%p|n*n<2=0|mod n p>0=n%(p+1)|r<-div n p=r+p*r%2
(%2)

Implements the recursive definition directly, with an auxiliary variable p that counts up to search for potential prime factors, starting from 2. The last line is the main function, which plugs p=2 to the binary function defined in the first line.

The function checks each case in turn:

  • If n*n<2, then n is one of -1,0,1, and the result is 0.
  • If n is not a multiple of p, then increment p and continue.
  • Otherwise, express n=p*r, and by the "derivative" property, the result is r*a(p)+p*a(r), which simplifies to r+p*a(r) because p is prime.

The last case saves bytes by binding r in a guard, which also avoids the 1>0 for the boilerplate otherwise. If r could be bound earlier, the second condition mod n p>0 could be checked as r*p==n, which is 3 bytes shorter, but I don't see how to do that.

Source Link
xnor
xnor
  • 146.6k
  • 26
  • 279
  • 652

Haskell, 59 bytes

n%p|n*n<2=0|mod n p>0=n%(p+1)|r<-div n p=r+p*r%2
(%2)

The last line is the main function, with the first line defining an auxiliary function. Implements the recursive definition directly, with an auxiliary variable p that counts up to search for potential prime factors, starting from 2.

The function checks each case in turn:

  • If n*n<2, then n is one of -1,0,1, and the result is 0.
  • If n is not a multiple of p, then increment p and recurse.
  • Otherwise, express n=p*r, and by the "derivative" property, the result is r*a(p)+p*a(r), which simplifies to r+p*a(r) because p is prime.

For the last case, we save bytes by binding r in a guard, while at the same time saving on 1>0 for the boilerplate otherwise. If r could be bound earlier, the second condition mod n p>0 could be checked as r*p==n, which is 3 bytes shorter, but I don't know how to do that.