OFFSET
1,1
COMMENTS
This is similar to but distinct from the even-indexed terms of A060796, with a(n) differing from A060796(2n) at n=7, 10, 11, 12, 13 and 16 (with A060796(36) unavailable for comparison). A260075 is the analog by splitting the first 3n primes into 3 equal-sized sets (but not by giving the smallest product larger than the cube root of the corresponding primorial). The percentages by which a(n) exceeds the square root of the (2n)-th primorial are 22.5, 3.51, 5.03, 0.660, 1.13, 0.347, 0.136, 1.82*10^(-3), 8.54*10^(-3), 6.21*10^(-3), 9.28*10^(-4), 1.84*10^(-4), 1.71*10^(-4), 1.31*10^(-5), 1.94*10^(-6), 5.62*10^(-8), 2.93*10^(-7) and 4.50*10^(-8).
The below PARI program functions by checking for each set of n primes through the (2n-1)-st whether either its product or its product's cofactor in the (2n)-th primorial gives an improvement.
LINKS
James G. Merickel, Table of n, a(n) for n = 1..18
EXAMPLE
If the primes 2 through 13 are split into equal-sized sets, the larger product of the two is not less than 182. This is the value for {2, 7, 13}, so a(3)=182.
PROG
(PARI)
{
p=vector(50, n, prime(n)); i=1;
while(1,
P=prod(j=1, 2*i, p[j]); Q=sqrt(P); r=10^1000;
m=prod(j=1, i, p[j]); a=vectorsmall(2*i-1);
for(j=1, i, a[j]=1); b=vectorsmall(2*i-1);
for(j=i, 2*i-1, b[j]=1); while(1,
if(m>Q, if(m<r, r=m), if(P/m<r, r=P/m));
if(a==b,
print1(r" "); break(),
k=1; if(a[1]==0,
while(a[k]==0, k++); c=1;
while(a[k+1], m*=p[c]/p[k]; a[c]=1; a[k]=0; c++; k++);
a[k]=0; a[k+1]=1; m*=p[k+1]/p[k],
while(a[k+1], k++); a[k]=0; a[k+1]=1; m*=p[k+1]/p[k])));
i++)
}
CROSSREFS
KEYWORD
nonn
AUTHOR
James G. Merickel, Jul 15 2015
STATUS
approved