OFFSET
1,1
COMMENTS
The rudimentary method employed by the PARI program below reaches the limit of its usefulness here. Contrast it with the method required for A259350, which is over 4.5 orders of magnitude faster than the analog of this (and may still be some distance best).
a(1)=A093550(6) (that sequence's 5th term, with offset 2). The program arbitrarily makes use of this knowledge, but will run (slower) without it.
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..10000
FORMULA
EXAMPLE
1990586013 = 3*13*29*67*109*241,
1990586014 = 2*23*37*43*59*461, and
1990586015 = 5*11*17*19*89*1259; and no smaller trio of this kind exists, making the middle value a(1).
PROG
(PARI)
{
\\Program initialized with known a(1).\\
\\The purpose of vector s and value u\\
\\is to skip bad values modulo 36.\\
k=1990586014; s=[4, 4, 8, 8, 8, 4]; u=1;
while(1,
if(issquarefree(k),
if(issquarefree(k-1),
if(issquarefree(k+1),
if(omega(k)==6,
if(omega(k-1)==6,
if(omega(k+1)==6,
print1(k" ")))))));
k+=s[u]; if(u==6, u=1, u++))
}
CROSSREFS
KEYWORD
nonn
AUTHOR
James G. Merickel, Jun 24 2015
STATUS
approved