cp's OEIS Frontend

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

%I A260075 #34 Jan 28 2016 18:15:38
%S A260075 5,35,627,20553,859066,48993082,3441790495,287535325407,
%T A260075 28839054633794,3161858853009549,416108939893639594,
%U A260075 60850811089314245258,9874934149007840709407,1754123227439445139773155
%N A260075 Where the first 3n primes are partitioned into 3 sets of cardinality n, the minimum of the largest product.
%C A260075 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.
%C A260075 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.
%C A260075 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).
%H A260075 James G. Merickel, <a href="/A260075/b260075.txt">Table of n, a(n) for n = 1..14</a>
%e A260075 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).
%o A260075 (PARI)
%o A260075 {
%o A260075 p=vector(60,n,prime(n));i=1;while(1,
%o A260075   a=vectorsmall(3*i);for(j=1,i,a[j]=1);
%o A260075   n=prod(j=1,i,p[j]);r=10^1000;
%o A260075   P=prod(j=1,3*i,p[j]);Q=P^(1/3);
%o A260075   b=vectorsmall(3*i);
%o A260075   for(j=2*i+1,3*i,b[j]=1);
%o A260075   while(1,
%o A260075     if(n<r && n>Q,
%o A260075         R=P/n;c=vector(2*i);k=1;
%o A260075         for(j=1,2*i,while(a[k],k++);
%o A260075           c[j]=k;k++);
%o A260075         d=vectorsmall(2*i);for(j=1,i,d[j]=1);
%o A260075         e=vectorsmall(2*i);for(j=i+1,2*i,e[j]=1);
%o A260075         S=prod(j=1,i,p[c[j]]);
%o A260075         while(1,
%o A260075           if(S<n && R/S<n,r=n;break());
%o A260075           if(d==e,
%o A260075             break(),
%o A260075             k=1;if(d[1]==0,
%o A260075               while(d[k]==0,k++);z=1;
%o A260075               while(d[k+1],S*=p[c[z]]/p[c[k]];
%o A260075                 d[z]=1;d[k]=0;z++;k++);
%o A260075               d[k+1]=1;d[k]=0;S*=p[c[k+1]]/p[c[k]],
%o A260075               while(d[k+1],k++);d[k]=0;d[k+1]=1;
%o A260075               S*=p[c[k+1]]/p[c[k]]))));
%o A260075   if(a==b,
%o A260075     print1(r" ");break(),
%o A260075     k=1;if(a[1]==0,
%o A260075       while(a[k]==0,k++);z=1;
%o A260075       while(a[k+1],n*=p[z]/p[k];a[z]=1;a[k]=0;z++;k++);
%o A260075       n*=p[k+1]/p[k];a[k]=0;a[k+1]=1,
%o A260075       while(a[k+1],k++);a[k]=0;a[k+1]=1;n*=p[k+1]/p[k])));i++)
%o A260075 }
%Y A260075 Cf. A260079.
%K A260075 nonn
%O A260075 1,1
%A A260075 _James G. Merickel_, Jul 14 2015