A260075 Where the first 3n primes are partitioned into 3 sets of cardinality n, the minimum of the largest product.

5, 35, 627, 20553, 859066, 48993082, 3441790495, 287535325407, 28839054633794, 3161858853009549, 416108939893639594, 60850811089314245258, 9874934149007840709407, 1754123227439445139773155

Offset: 1

Author

James G. Merickel, Jul 14 2015

Keywords

• nonn

Comments

Distinct from the smallest product of n primes from the first 3n that is larger than the cube root of the (3n)-th primorial. Both this sequence and that (hypothetical) one are analogs of A260079.

The below PARI program runs through each product of n primes up to the (3n)-th, testing it for whether it is both greater than the cube root of the (3n)-th primorial and less than the smallest result to that point; and, if these limitations are met, it then goes on to determine whether or not the remaining primes can be split into n-cardinality halves with products both less than it.

The percentages by which a(n) exceeds the (3n)-th primorial's cube root are 60.9, 12.6, 3.38, 5.37, 1.02, 0.0883, 0.0340, 5.18*10^(-3), 5.01*10^(-4), 1.68*10^(-4), 2.37*10^(-5), 2.06*10^(-5), 3.87*10^(-5) and 1.14*10^(-6).

Examples

			There are 15 distinct ways to break a set of six elements into three of cardinality 2 each. Among these for the first 6 primes, the partition {{2,13}, {3,11}, {5,7}} can be readily seen to give smallest possible maximum product of 35. So a(2)=35 (and with the convention that the product of a set consisting of one number is that number, a(1)=5 trivially).
		

Links

• James G. Merickel, Table of n, a(n) for n = 1..14

Crossrefs

Cf. A260079.

Programs

PARI

{
p=vector(60,n,prime(n));i=1;while(1,
  a=vectorsmall(3*i);for(j=1,i,a[j]=1);
  n=prod(j=1,i,p[j]);r=10^1000;
  P=prod(j=1,3*i,p[j]);Q=P^(1/3);
  b=vectorsmall(3*i);
  for(j=2*i+1,3*i,b[j]=1);
  while(1,
    if(nQ,
        R=P/n;c=vector(2*i);k=1;
        for(j=1,2*i,while(a[k],k++);
          c[j]=k;k++);
        d=vectorsmall(2*i);for(j=1,i,d[j]=1);
        e=vectorsmall(2*i);for(j=i+1,2*i,e[j]=1);
        S=prod(j=1,i,p[c[j]]);
        while(1,
          if(S