OFFSET
1,1
COMMENTS
A subsequence of A169834. A093550(7)=a(1), that sequence with offset 2 (so actually its 6th term) holding first terms of sequences of this kind.
Other than a(4)=366230785766 and a(18)=905173650094 (with minimax prime factor 1867 for it and its neighbors), the terms were initially discovered by increasing value of the trios' smallest large prime factors. An exhaustive search running multiple (suitably modified) copies of a pre-acceptance PARI program that disposed of fails in a somewhat efficient way and ran about an order of magnitude faster than the analog of the simple program at A259349 required about 1000 window-hours to produce the list given (adding two terms, including one that was unachievable by the increasing-minimax-prime method). Then the much faster program--15 minutes in just one PARI window--shown was developed and edited in here in its place. By specifying the 4 largest prime factors secondary to setting the product of the smallest 3 such that this is at least 627--as must be true for one of 3 relatively prime sphenic numbers--a speedup of over 3.5 orders of magnitude more (over the single order of magnitude that the replaced program managed, for a total of about 10^4.5 in time ratio over the program used for 6 primes) was achieved.
Note: The PARI program avoids duplicates but does not order terms.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..2510
EXAMPLE
41704979953 = 7*13*29*41*47*59*139,
41704979954 = 2*11*23*31*83*103*311, and
41704979955 = 3*5*17*19*109*157*503; and no smaller such trio exists, so that a(1)=41704979954.
PROG
(PARI)
{
\\Program runs for arbitrary B.\\
B=10^12; N=primepi(B/(627*17*19*23));
p=vector(N, n, prime(n));
in=primepi((B/210)^(1/3));
P=prod(i=1, 27, p[i]); Q=prod(i=28, in, p[i]);
v=28; d=[[1, 2], [-1, 1], [-2, -1]]; i3=6;
while(6*p[i3]^5<B,
k1=p[i3];
for(i2=2, i3-1,
k1*=p[i2];
for(i1=1, i2-1,
k1*=p[i1];
if(k1>626,
if(k1*p[i3+1]*p[i3+2]*p[i3+3]*p[i3+4]<B,
if(6*p[v-1]<k1,
while(6*p[v]<k1, Q/=p[v]; P*=p[v]; v++),
while(6*p[v-1]>=k1, v--; Q*=p[v]; P/=p[v]));
r=(B\k1)^(1/4); j1=i3+1;
while(p[j1]<r,
k2=p[j1]*k1; s(B\k2)^(1/3); j2=j1+1;
while(p[j2]<s,
k3=p[j2]*k2; t=sqrt(B\k3); j3=j2+1;
while(p[j3]<t,
k4=k3*p[j3]; j4=j3+1; u=B\k4;
while(p[j4]<=u,
K=k4*p[j4];
a=K+d[K%4][1]; b=K+d[K%4][2];
a1=gcd(P, a); y1=omega(a1);
if(y1>2,
f=1; if(y1==3, if(a1>j1, f=0));
if(f,
b1=gcd(P, b); z1=omega(b1);
if(z1>2,
if(z1==3, if(b1>j1, f=0));
if(f,
a2=a/a1;
if(gcd(a1, a2)==1,
b2=b/b1;
if(gcd(b1, b2)==1,
a21=gcd(a2, Q); a22=a2/a21;
if(gcd(a21, a22)==1,
y=y1+omega(a21);
if(y>4,
if(y<8,
b21=gcd(Q, b2); b22=b2/b21;
if(gcd(b21, b22)==1,
z=z1+omega(b21);
if(z>4,
if(z<8,
if(y+omega(a22)==7,
if(z+omega(b22)==7,
f1=factor(a1);
if(f1[1, 1]*f1[2, 1]*f1[3, 1]<k1,
f2=factor(b1);
if(f2[1, 1]*f2[2, 1]*f2[3, 1]<k1,
print1((K\4)*4+2" ")
))))))))))))))));
j4++);
j3++);
j2++);
j1++)));
k1/=p[i1]);
k1/=p[i2]);
i3++)
}
CROSSREFS
KEYWORD
nonn
AUTHOR
James G. Merickel, Jun 24 2015
STATUS
approved