OFFSET
1,1
COMMENTS
The only consecutive brilliant numbers are {9, 10} and {14, 15}; and for m > 14 there are no brilliant constellations of the form {m, m+(2k+1)} or equivalently {n, 2k+m+1} with k >= 0. Proof: One of m and 2k+m+1 will be even. And there are no even brilliant numbers > 14 since they must have the form 2*p where p is a prime having only one digit.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
EXAMPLE
a(3) = 779 because 779=19*41 and 781=11*71.
CROSSREFS
KEYWORD
base,easy,nonn
AUTHOR
Jason Earls, Jun 03 2003
STATUS
approved