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 A260075 Where the first 3n primes are partitioned into 3 sets of cardinality n, the minimum of the largest product. 3
 5, 35, 627, 20553, 859066, 48993082, 3441790495, 287535325407, 28839054633794, 3161858853009549, 416108939893639594, 60850811089314245258, 9874934149007840709407, 1754123227439445139773155 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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). LINKS James G. Merickel, Table of n, a(n) for n = 1..14 EXAMPLE 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). PROG (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

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Last modified July 23 13:58 EDT 2019. Contains 325254 sequences. (Running on oeis4.)