A255597 Upper bound on the number of different Euler diagrams for n classes.

1, 1, 3, 29, 1667, 3254781, 10650037396483, 113423713055347294030815229, 12864938683278671740537145090971257103576706009186307

Offset: 0

Author

Maurizio De Leo, Feb 27 2015

Keywords

• nonn

Comments

Obtained via an iterative method. Each new class added to an existing diagram must create at least a new zone, and at most a number of new zones equal to the existing zones.

Examples

			For n=3 (3 different classes) there are 29 possible Euler diagrams that do not reduce to smaller cases. Of these 11 are in fact repetitions and need to be eliminated to perfect the upper bound.
		

Links

• Mathematics Stack Exchange, How many Euler diagrams with n sets exist?.
• Wikipedia, Euler diagram.

Crossrefs

This sequence is linked to A007018 by the binomial transform: A007018(n) = Sum_{k=0..n} C(n,k)*a(k).

Programs

C

#include 
#include 
#include 
#define MAXCLU 7
#define MAXZONE 256
long long combi(int n, int k){
    if (nn-k?n-k:k;
    int j=1;
    for(;j<=k;j++,n--){
        if(n%j==0){
            ans*=n/j;
        }else if(ans%j==0){
            ans=ans/j*n;
        }
        else{
            ans=(ans*n)/j;
        }
    }
    return ans;
}
int main(){
    long long  a[MAXCLU][MAXZONE];
    long long sum[MAXCLU];
    int j,k,i;
    for (j=0;j
				

Formula

a(n) = Sum_{k>=1} e(n,k), where k is the number of zones, and the elements e(n,k) are defined recursively as: e(0,1) = 1; e(n,k) = Sum_{c=1..k-1} binomial(c,k-c)*e(n-1,c).

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*A007018(k). - Jason Yuen, Mar 01 2025

Extensions

More terms from Jason Yuen, Mar 01 2025