A104461 Number of instances of nonprimes m in Pythagorean triples x,y,z such that x^2 + y^2 = z^2. Except for 1, the number of instances of composite numbers m in Pythagorean triples.

0, 1, 1, 2, 2, 2, 4, 1, 5, 3, 2, 5, 4, 1, 7, 4, 2, 3, 4, 5, 4, 4, 2, 5, 7, 1, 5, 8, 4, 4, 8, 1, 10, 2, 4, 5, 5, 3, 5, 7, 4, 2, 14, 1, 7, 5, 8, 4, 5, 4, 5, 12, 2, 9, 4, 4, 5, 11, 4, 2, 13, 8, 1, 5, 7, 8, 5, 4, 4, 1, 5, 13, 2, 7, 9, 5, 8, 14, 2, 10, 5, 5, 10, 4, 5, 5, 8, 1, 5, 23, 2, 2, 5, 4, 6, 7, 6, 4, 8, 13

Offset: 1

Author

Cino Hilliard, Apr 18 2005

Keywords

• easy
• nonn

Comments

The PARI script is direct and very fast for m = x,y values but slows in the trial routine for m = z. We save some time for m even allowing the testing of only even values of y.

Examples

			For m=30 there are 5 Pythagorean triples that have a 30:
  30, 224, 226
  30,  72,  78
  30,  40,  50
  30,  16,  34
  18,  24,  30
		

Crossrefs

Cf. A046081, A088978.

Programs

PARI

\\ instances of m in Pythagorean triples using a direct method for x,y
pythm3(m) = { local(m2,ln,j,j2=0,d,d2,q2,q,a,b,x,x1,x2,xx,y,y2,z,c,c2,r,f,str,stp); d=divisors(m^2); /* get the divisors of m^2 */ ln=length(d)-1; d2=q2=vector(ln); m2=m^2; if(m%2,r=1,r=0); for(j=1,ln, /* save only the both even r=0, both odd r=1 */ if(d[j]%2==r, if(m2/d[j]%2==r, j2++; d2[j2]=d[j]; q2[j2]=m2/d[j]; /* save m/factor to solve (z-y)(z+y) = m^2 */ ) ) ); x2=y2 = vector(20); for(j=1,j2, z=(d2[j] + q2[j])/2; y= z - d2[j]; if(y>0, c++; ) ); if(m%2==0,start=2;step=2,start=1;step=1); forstep(y=start,m-1,step, /* esolve when z is m */ x1 = (m2-y^2); if(issquare(x1), c2++; x2[c2]=floor(sqrt(x1)); /* save to later mask dupes */ y2[c2]=y; ) ); for(x=1,c2, /* mask the dupes routine */ for(y=x,c2, if(x2[x]==y2[y], ) ) ); return(c+c2/2) /* print total */}
for(k=1,400,if(isprime(k)==0,print1(pythm3(k)", ")))

Formula

Consider Pythagorean triples x^2 + y^2 = z^2. We seek to find the total number of instances of an integer m being x or y or z. The solution for x or y is straightforward by considering appropriate lesser and greater pairwise factors, L, G of m^2 in z^2 - y^2 = (z-y)(z+y) = m^2. Then solve for z and y with the relations, z-y = L z+y = G 2z = L+G, z = (L+G)/2 where L and G are both even if m is even or both odd if m is odd. The number of L factors < m is the number of instances of x or y. The count of instances z=m is solved by trial on x^2 = m^2 - y^2.