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COMMENTS
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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).
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EXAMPLE
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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).
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PROG
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(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(n<r && n>Q,
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<n && R/S<n, r=n; break());
if(d==e,
break(),
k=1; if(d[1]==0,
while(d[k]==0, k++); z=1;
while(d[k+1], S*=p[c[z]]/p[c[k]];
d[z]=1; d[k]=0; z++; k++);
d[k+1]=1; d[k]=0; S*=p[c[k+1]]/p[c[k]],
while(d[k+1], k++); d[k]=0; d[k+1]=1;
S*=p[c[k+1]]/p[c[k]]))));
if(a==b,
print1(r" "); break(),
k=1; if(a[1]==0,
while(a[k]==0, k++); z=1;
while(a[k+1], n*=p[z]/p[k]; a[z]=1; a[k]=0; z++; k++);
n*=p[k+1]/p[k]; a[k]=0; a[k+1]=1,
while(a[k+1], k++); a[k]=0; a[k+1]=1; n*=p[k+1]/p[k]))); i++)
}
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