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Quick update on the last post. I have shown a graph for the length of rhythmic canons mod 2 with motive $\{0,1,k\}$, and we have seen that the dependence of the length with $k$ is everything but simple, with particular cases when $k=2^p$ or $k=2^p+1$. I wanted to check what happens for values of $k$ higher than 41, but it turns out that the computation of the length is now very slow. I managed to get the lengths for $k \in \{42,43,45,47,49\}$, and these are the red points in the updated graph below. I am waiting for other values of $k$, but the computation for $k=42$ (haha…) took nearly two weeks, so this won’t be for tomorrow.

What are the green points by the way ? Well, I’m sure you have noticed that the previous code can easily be adapted to get rhythmic canons for any modulus, and not only 2. Here is an improved C program for calculating the length L (and entries) of a canon mod P with motive $\{0,1,k\}$. Notice that you will get the version of the canon where entries may be repeated, and that you may “unravel the modulus” of any entry at position $x$ by considering another entry at position $x+L$. The green points in the graph above are the length of the canons with motive $\{0,1,k\}$ modulus 3.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char *argv[])
{
int N,P;
int* U;
int i,mark=0,flag=0;
unsigned long long L,entry=0;
FILE* output_file=NULL;

if(argc<3) {
printf("Length of the motive is required !\n");
exit(1);
}
N = atoi(argv[1]);
P = atoi(argv[2]);
if(argc==4)
output_file=fopen(argv[4],"w");
if (output_file!=NULL)
fprintf(output_file,"Entries for canon with motive {0,1,%d}\n",N);

U = calloc(N+1,sizeof(int));
if (U==NULL) {
printf("Error during memory allocation...exiting now !\n");
exit(1);
}

L=N+1;

while(flag==0) {

L+=mark;
entry+=mark;
if (output_file!=NULL)
fprintf(output_file,"%llu\n",entry);

U[0]= (U[0]+1) % P;
U[1]= (U[1]+1) % P;
U[N]= (U[N]+1) % P;

flag=1;
for (mark=0; mark<N+1; mark++) {
if (U[mark]!=1) {
flag=0;
break;
}
}

for (i=mark; i<N+1; i++)
U[i-mark]=U[i];
for (i=(N+1-mark); i<N+1; i++)
U[i]=0;
}
printf("Final length of canon with motive {0,1,%d}: %llu \n",N,L);
if (output_file!=NULL)
fclose(output_file);
free(U);
}


Interestingly, it seems that any canon with motive $\{0,1,3^p\}$ has length $3(3^{2p}-1)$, and any canon with motive $\{0,1,3^p+1\}$ has length $3(3^{2p}+3^p+1)$. I am guessing that for any prime modulus, a similar relation exists…

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